Quadratic Equations - Class 10 Mathematics - Chapter 4 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Quadratic Equations | NCERT | Mathematics | Class 10
By increasing the list price of a book by ₹10, a person can buy 20 books less for ₹1200. The original price of the book was _____.
A) ₹0
B) ₹20
C) ₹30
D) ₹24
The problem provides the scenario where increasing the list price of a book by ₹10 results in buying 20 fewer books for ₹1200. We are asked to find the original price of the book.
Let's denote the original price of the book as $ x $. Hence, after the price increase, the new price of the book becomes $ x + 10. $
The number of books that can be bought initially with ₹1200 is given by $ \frac{1200}{x},$ and the number of books that can be bought after the price increase is $ \frac{1200}{x + 10}. $
According to the problem, the difference in the number of books bought before and after the price increase is 20 books, leading to the equation: $$ \frac{1200}{x} - \frac{1200}{x + 10} = 20 $$
We can simplify and solve this equation by first clearing the denominators: $$ 1200(x + 10) - 1200x = 20x(x + 10) $$ $$ 12000 = 20x^2 + 200x $$
Dividing all terms by 20 to simplify: $$ 600 = x^2 + 10x $$
Rearranging into a standard quadratic equation form: $$ x^2 + 10x - 600 = 0 $$
Solving this quadratic equation using the quadratic formula where $ a = 1 $, $ b = 10 $, and $ c = -600 $: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ $$ x = \frac{-10 \pm \sqrt{100 + 2400}}{2} $$ $$ x = \frac{-10 \pm \sqrt{2500}}{2} $$ $$ x = \frac{-10 \pm 50}{2} $$
We have two potential solutions for $ x $: $$ x = \frac{40}{2} = 20 \quad \text{and} \quad x = \frac{-60}{2} = -30 $$
Rejecting the negative value, $ x = -30 $, as a price cannot be negative, the original price of the book is therefore ₹20. Hence, the correct answer is:
B) ₹20
The area of a rectangular plot is $528 \mathrm{~m}^{2}$. The length of the plot (in metres) is one more than twice its breadth. Find the length and breadth of the plot.
A) $30 \mathrm{~m}$ and $10 \mathrm{~m}$
B) $35 \mathrm{~m}$ and $12 \mathrm{~m}$
C) $34 \mathrm{~m}$ and $15 \mathrm{~m}$
D) $33 \mathrm{~m}$ and $16 \mathrm{~m}$
The correct option is D) $33 \text{ m}$ and $16 \text{ m}$
Assume the breadth of the rectangular plot to be $x$ meters. Then, the length of the plot, being one more than twice its breadth, is given by: $$ \text{Length} = 2x + 1 \text{ meters} $$ According to the problem, the area of the rectangle is $528 \text{ m}^2$. We express the area using the formula for the area of a rectangle: $$ \text{Area} = \text{Length} \times \text{Breadth} = (2x + 1) \times x = 528 $$ Expanding and setting up the equation we get: $$ 2x^2 + x = 528 $$ Rearrange to form a quadratic equation: $$ 2x^2 + x - 528 = 0 $$
We factorize the quadratic equation: $$ 2x^2 - 32x + 33x - 528 = 0 $$ Group the terms: $$ (2x^2 - 32x) + (33x - 528) = 0 $$ Factoring out the common terms: $$ 2x(x - 16) + 33(x - 16) = 0 $$ Factor further: $$ (2x + 33)(x - 16) = 0 $$
Setting each factor to zero, we get: $$ 2x + 33 = 0 \quad \text{or} \quad x - 16 = 0 $$ Solving these, we find: $$ x = -\frac{33}{2} \quad \text{(not possible since breadth cannot be negative)} \quad \text{or} \quad x = 16 $$
Therefore, the breadth is $x = 16$ m. Substituting this into the length equation: $$ \text{Length} = 2(16) + 1 = 33 \text{ m} $$
Thus, the length is $33$ m and the breadth is $16$ m.
If Abhilash was younger by 2 years than what he really is, then the square of his age (in years) would have been 68 more than 8 times his actual age. What is his age now?
A) 16 years
B) 50 years
C) 20 years
D) 29 years
The correct option is A) 16 years
Let us denote Abhilash's current age as $x$ years. If he were 2 years younger, his age would be $(x-2)$ years. The problem states that at this younger age, the square of his age would be 68 units greater than 8 times his current age. Thus, the equation formulated from this condition is: $$ (x-2)^2 = 8x + 68 $$
Expanding and simplifying this equation gives: $$ \begin{align*} (x - 2)^2 & = 8x + 68 \ x^2 - 4x + 4 & = 8x + 68 \ x^2 - 12x - 64 & = 0 \end{align*} $$
Factoring the quadratic equation: $$ \begin{align*} x^2 - 16x + 4x - 64 & = 0 \ x(x - 16) + 4(x - 16) & = 0 \ (x + 4)(x - 16) & = 0 \end{align*} $$
By the Zero Product Property, $x + 4 = 0$ or $x - 16 = 0$. Solving these gives: $$ x = -4 \quad \text{or} \quad x = 16 $$
Since age cannot be negative, $x=-4$ is not feasible. Therefore, Abhilash's current age is $x = 16$ years.
Find $x$.
$x^{2} - 9 = 72 $
A. 7
B. 9
C. 8
D. 10
Given the equation:
$$ x^{2} - 9 = 72 $$
To isolate $x^2$, add 9 to both sides:
$$ x^{2} = 72 + 9 $$
$$ x^{2} = 81 $$
To find the value of $x$, take the square root of both sides. Remember, the square root has two values, positive and negative:
$$ x = \pm\sqrt{81} $$
$$ x = \pm9 $$
Both $9$ and $-9$ are solutions, but we should check which one fits with the given options. The available choices are:
A. 7
B. 9
C. 8
D. 10
Among the given options, 9 (option $\mathbf{B}$) is the correct answer.
Thus, $\boxed{9}$ is the solution.
If $\alpha$ and $\beta$ are the roots of the equation $a x^{2} + b x + c = 0$, where $(a, b, c \in \mathbb{R})$, then $\left(1 + a + a^{2}\right) \left(1 + \beta + \beta^{2}\right)$ is:
A $<0$
B $>0$
C $=0$
D None of these
The correct option is B
Firstly, using Vieta's formulas, we know: $$ \alpha + \beta = -\frac{b}{a} $$ $$ \alpha \beta = \frac{c}{a} $$
We aim to compute the expression:
$$
\left(1 + \alpha + \alpha^2\right)\left(1 + \beta + \beta^2\right)
$$
Expanding and substituting terms, we get:
$$
\begin{align*}
&= 1 + (\alpha + \beta) + (\alpha^2 + \beta^2) + \alpha \beta + \alpha \beta(\alpha + \beta) + \alpha^2 \beta^2 \
&= 1 + (-\frac{b}{a}) + (\alpha + \beta)^2 - \alpha \beta + \alpha \beta(-\frac{b}{a}) + (\alpha \beta)^2\
&= 1 - \frac{b}{a} + \frac{b^2}{a^2} - \frac{c}{a} - \frac{bc}{a^2} + \frac{c^2}{a^2} \
&= \frac{\left[a^2 + b^2 + c^2 - ab - bc - ca\right]}{a^2} \
&= \frac{\left[(a+b)^2 + (b-c)^2 + (c-a)^2\right]}{2a^2}
\end{align*}
$$
Since the expression is a sum of squared terms divided by $2a^2$, and squares are non-negative and the denominator $2a^2$ is positive for non-zero $a$, the entire expression is non-negative and non-zero. Therefore, it is definitely positive: $0 $
So, the correct answer is B.
$\sqrt{3}x - y = 2$ and $\sqrt{6}x - \sqrt{2}y = 2$ are lines.
A) coincident
B) parallel
C) intersecting
To determine the relationship between the two lines given by the equations $$\sqrt{3}x - y = 2$$ and $$\sqrt{6}x - \sqrt{2}y = 2$$, we need to check for their slopes and intercepts.
Step 1: Convert the given equations into the standard line equation form, $ax + by + c = 0$.
For $\sqrt{3}x - y = 2$, it can be converted to: $$ \sqrt{3}x - y - 2 = 0 $$
For $\sqrt{6}x - \sqrt{2}y = 2$, it can be converted to: $$ \sqrt{6}x - \sqrt{2}y - 2 = 0 $$
Step 2: Use the proportionality condition for parallel lines, which requires: $$ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} $$
Compare coefficients from both line equations:
Coefficient for $x$: $$ \frac{\sqrt{3}}{\sqrt{6}} = \frac{1}{\sqrt{2}} $$
Coefficient for $y$: $$ \frac{-1}{-\sqrt{2}} = \frac{1}{\sqrt{2}} $$
Constant term: $$ \frac{-2}{-2} = 1 $$
Assembly: Since $\frac{\sqrt{3}}{\sqrt{6}} = \frac{-1}{-\sqrt{2}} = \frac{1}{\sqrt{2}}$ but the ratios involving the constants do not match the slope ratios, the lines are parallel rather than coincident.
Hence, the correct answer is (B) parallel.
A man invested Rs 45,000 in 15% Rs 100 shares quoted at Rs 125. When the market value of these shares rose to Rs 140, he sold some shares, just enough to raise Rs 8400. Calculate the: (i) Number of shares he still holds; (ii) Dividend due to him on these remaining shares. [3 MARKS]
Each subpart: 1.5 Marks
(i) Initially, the number of shares purchased is calculated by dividing the total investment by the price per share: $$ \text{Number of shares} = \frac{45000}{125} = 360 \text{ shares} $$
He sold enough shares to raise Rs 8400, when the market price changed to Rs 140. $$ \text{Number of shares sold} = \frac{8400}{140} = 60 \text{ shares} $$ Thus, the number of shares he still holds is: $$ 360 - 60 = 300 \text{ shares} $$
(ii) The dividend per share is 15% of the face value Rs 100, which is: $$ \frac{15}{100} \times 100 = 15 \text{ Rs per share} $$ Thus, the total dividend on 300 shares is: $$ 15 \times 300 = 4500 \text{ Rs} $$
Without solving, examine the nature of roots of the equation.
$$ 4x^{2} - 4x + 1 = 0 $$
The provided quadratic equation is: $$ 4x^2 - 4x + 1 = 0 $$
From this, we identify the coefficients: $$ a = 4, \quad b = -4, \quad c = 1 $$
We determine the discriminant ($D$) using the formula: $$ D = b^2 - 4ac $$
Substituting the values, we calculate: $$ D = (-4)^2 - 4 \cdot 4 \cdot 1 = 16 - 16 = 0 $$
Since $D=0$, the roots of the quadratic equation are real and equal.
Which of the following equations has the roots 7 and 8?
A) $x^2 - x + 19 = 0$
B) $x^2 - 19x + 1 = 0$
C) $x^2 - 17x + 2 = 0$
D) $x^2 - 15x + 56 = 0$
Solution:
To determine the correct equation, we use the sum and product of the roots of a quadratic equation. Given roots are $7$ and $8$.
The sum of the roots, $(A+B)$, and the product of the roots, $(AB)$, relate to the coefficients in the standard form of a quadratic equation: $$ x^2 - (A+B)x + AB = 0 $$ Here, substituting $A = 7$ and $B = 8$: $$ x^2 - (7+8)x + (7 \times 8) = 0 $$ This simplifies to: $$ x^2 - 15x + 56 = 0 $$
Matching this with the given options, we find that option D) $x^2 - 15x + 56 = 0$ matches the equation derived from the roots 7 and 8. Thus, the correct answer is D.
If $a, \beta$ are the roots of the equation $x^{2} - ax + b = 0$, and the roots of the equation $bx^{2} - 4x + 4 = 0$ are $a + \frac{\beta^{2}}{a}, \beta + \frac{a^{2}}{\beta}$, then $a + b$ can be -
A. 0
B. 1
C. 2
D. 3
To solve the equations provided, wherein $a$ and $\beta$ are the roots of the equation $$ x^{2} - ax + b = 0, $$ and the roots of another equation $$ bx^{2} - 4x + 4 = 0 $$ are given as $a + \frac{\beta^{2}}{a}, \beta + \frac{a^{2}}{\beta}$, we need to determine the possible values for $a + b$.
From the first equation, we know the product and sum of the roots based on Vieta's formulas: $$ a + \beta = a \quad \text{and} \quad a\beta = b. $$
Substitute the expressions for the roots into the second equation:
- For the root $a + \frac{\beta^{2}}{a}$, $$ a + \frac{\beta^{2}}{a} = \frac{a^2 + \beta^{2}}{a} = \frac{a^2 + (\beta^2)}{a}. $$ Since $\beta = a - \beta$, this expression simplifies to $\frac{a^2 + a^2 - 2a\beta}{a}$, leading to: $$ a + \frac{\beta^2}{a} = \frac{a^2 - 2b}{a} = a - \frac{2b}{a}. $$
- Similarly, for the root $\beta + \frac{a^{2}}{\beta}$: $$ \beta + \frac{a^2}{\beta} = \frac{\beta^2 + a^2}{\beta} = \frac{a^2 + a^2 - 2a\beta}{\beta} = a - \frac{2b}{\beta}. $$
The resulting quadratic equation with roots $a - \frac{2b}{a}$ and $a - \frac{2b}{\beta}$ is: $$ x^2 - \left(a - \frac{2b}{a} + a - \frac{2b}{\beta}\right)x + \left(a - \frac{2b}{a}\right)\left(a - \frac{2b}{\beta}\right) = 0. $$
Expanding and simplifying this form, and comparing it to $bx^{2} - 4x + 4 = 0$, equate the coefficients: $$ b = 1 \quad \text{and} \quad a\left(a^{2} - 2b\right) = 4, \quad \text{and} \quad \left(a^{2} - 2b\right)^2 = b^2. $$
Solving these equations under the given conditions yields the possible values: $$ \begin{array}{l} a^2-2b = \pm 2 \ a(a^2-2b) = 4 \ \Rightarrow a = 2, b = 1 \Rightarrow 2 + 1 = 3 \ \text{or} \ a = -2, b = 3 \Rightarrow -2 + 3 = 1. \end{array} $$
Thus, the possible values for $a + b$ are 1 and 3. Therefore, the correct answers are:
- B. 1
- D. 3
If the two equations $x^{3} + 3px^{2} + 3qx + r = 0$ and $x^{2} + 2px + q = 0$ have a common root, then the value of $4(p^{2} - q)(q^{2} - pr)$ is
A $(pq + r)^{2}$
B $(pq - r)^{2}$
C $(p - qr)^{2}$
D $(p + qr)^{2}$
Here's the reframed solution:
Solution
The correct option is B - $(pq - r)^{2}$. Let's consider $a$ to be the common root for both given equations:
- The equation $x^{3} + 3px^{2} + 3qx + r = 0$ provides: $$ a^{3} + 3pa^{2} + 3qa + r = 0 \quad (1) $$
- The equation $x^{2} + 2px + q = 0$ gives: $$ a^{2} + 2pa + q = 0 \quad (2) $$
Multiplying equation (2) by -$a$, we derive: $$ -a(a^{2} + 2pa + q) = 0 \implies -a^{3} - 2pa^2 - aq = 0 \quad (3) $$
By adding equations (1) and (3), we cancel out $a^3$ and solve: $$ (3pa^2 - 2pa^2) + (3qa - aq) + r = 0 \implies pa^2 + 2qa + r = 0 \quad (4) $$
Notice that equations (2) and (4) share $a$ as a common root. We can now apply the common root condition for determinants:
$$ \begin{vmatrix} 1 & 2p & q \ p & 2q & r \end{vmatrix} $$
Expanding this determinant and simplifying using the common root condition: $$ (1 \times 2q - p \times 2p) \times (2p \times r - 2q \times q) = (q - pq)^2 $$
This simplifies further: $$ (2q - 2p^2)(2pr - 2q^2) = (pq - r)^2 $$
Multiplying through by 4 to clear the factor on each side: $$ 4(p^2 - q)(q^2 - pr) = (pq - r)^2 $$
Thus, the value of $4(p^{2} - q)(q^{2} - pr)$ is indeed $(pq - r)^2$, corresponding to option B.
Shriya and Vidya solved a quadratic equation. In solving it, Shriya made a mistake in the constant term and obtained the roots as $5, -3$ while Vidya made a mistake in the coefficient of $x$ and obtained the roots as $1, -3$. Which of the following are correct about the correct roots of the equation?
A) One of the correct roots is -1.
B) One of the correct roots is 1.
C) One of the correct roots is -3.
D) One of the correct roots is 3.
The correct options regarding the roots of the quadratic equation are:
A) One of the correct roots is -1.
D) One of the correct roots is 3.
Let's determine the correct quadratic equation based on the information given and solve for its roots:
Shriya's calculation had an error only in the constant term, which means that the sum of the roots she obtained can be considered accurate. The sum of roots obtained by Shriya is $5 + (-3) = 2$.
For a quadratic equation of the form $ax^2 + bx + c = 0$, the sum of the roots is given by: $$ \text{Sum of roots} = -\frac{b}{a} = 2 $$
Vidya's error was solely in the coefficient of $x$, which suggests that the product of the roots should be correct. Vidya's roots provide the product as $1 \times (-3) = -3$: $$ \text{Product of roots} = \frac{c}{a} = -3 $$
Thus, we can form the quadratic equation using these properties: $$ x^2 - 2x - 3 = 0 $$ This equation can be factored as: $$ x^2 - 3x + x - 3 = 0 \ x(x-3) + 1(x-3) = 0 \ (x+1)(x-3) = 0 $$
This factorization reveals the roots as: $$ x + 1 = 0 \rightarrow x = -1 \ x - 3 = 0 \rightarrow x = 3 $$
Thus, the correct roots of the equation are -1 and 3, confirming options A and D.
If a triangle is formed by any three tangents of the parabola $y^{2} = 4ax$ whose two of its vertices lie on $x^{2} = 4by$, then the third vertex lies on
A $(x-1)^{2} = 4ay$
B $x^{2} = 16ay$
C $x^{2} = 4by$
D $(x+1)^{2} = 4ay$
The correct option is C $x^{2} = 4by$.
Given a parabola with the equation $y^2 = 4ax$, we consider three tangent lines to it with the equations: $$ yt_1 = x + at_1^2, \quad yt_2 = x + at_2^2, \quad yt_3 = x + at_3^2. $$
The intersection points of these tangents, or the vertices of the triangle, are: $$ A(at_{1}t_{2}, a(t_{1}+t_{2})), \quad B(at_{2}t_{3}, a(t_{2}+t_{3})), \quad C(at_{1}t_{3}, a(t_{1}+t_{3})). $$
We know that vertices $A$ and $B$ lie on the curve $x^2=4by$. Applying this condition gives us the equations: $$ (a t_1 t_2)^2 = 4ab(t_1 + t_2), \quad (a t_2 t_3)^2 = 4ab(t_2+t_3). $$
To find a relation among $t_1$, $t_2$, and $t_3$, we use the division of these conditions: $$ \frac{t_1^2}{t_3^2} = \frac{t_1 + t_2}{t_2 + t_3} \Rightarrow t_2(t_1^2 - t_3^2) = t_1 t_3(t_3-t_1). $$ This leads us to conclude: $$ t_2 = -\frac{t_1 t_3}{t_1 + t_3}. $$
Subtracting these conditions, we simplify further: $$ a^2 t_2^2(t_1^2 - t_3^2) = 4ab(t_1 - t_3) \Rightarrow a^2\left(-\frac{t_1 t_3}{t_1 + t_3}\right)^2(t_1 + t_3) = 4ab. $$ Resolving gives: $$ (a t_1 t_3)^2 = 4a b (t_1 + t_3). $$
Thus, the coordinates of vertex $C$, given by $(a t_1 t_3, a(t_1 + t_3))$, must satisfy the equation $x^2 = 4by$. Therefore, vertex $C$ also lies on the curve $x^2 = 4by$, confirming that the correct answer is option C.
When we divide $x^{2} + 2x + 9$ by $x + 1$, we get the remainder as $\qquad$ A. 8.
According to the remainder theorem, when polynomial $P(x)$ is divided by $(x - a)$, the remainder is $P(a)$.
Given that $P(x) = x^2 + 2x + 9$, and it is divided by $x + 1$ which parallels $(x - (-1))$, we thus set $a = -1$.
Let's compute $P(-1)$: $$ P(-1) = (-1)^2 + 2(-1) + 9 = 1 - 2 + 9 = 8 $$
Thus, the remainder when $x^{2} + 2x + 9$ is divided by $x + 1$ is 8. Answer is A. 8.
What will be the quotient and remainder if $\left(6x^{3}+8x^{2}\right) \div 2x$?
A) Quotient $=3x^{2}+4x$ Remainder $=0$
B) Quotient $=3x+4$ Remainder $=0$
C) Remainder $=3x^{2}+4x^{3}$ Quotient $=12x$
D) Remainder $=3x^{2}+4x^{3}$ Quotient $=0$
To solve the division $\left(6x^3 + 8x^2\right) \div 2x$, we simplify by dividing each term in the polynomial individually by $2x$.
- Divide the term $6x^3$ by $2x$: $$ \frac{6x^3}{2x} = 3x^2 $$
- Divide the term $8x^2$ by $2x$: $$ \frac{8x^2}{2x} = 4x $$
Hence, the quotient is: $$ 3x^2 + 4x $$
Quotient: $3x^2 + 4x$ Remainder: 0
This corresponds to Option A:
- Quotient: $3x^2 + 4x$
- Remainder: 0
The sum and product of zeroes of a quadratic polynomial $ax^{2} + bx + c$ respectively are _ and _.
A) $-b/a$
B) $c/a$
C) $b/a$
D) $-c/a$
The solution to identifying the sum and product of zeroes for a quadratic polynomial of the form $$ ax^2 + bx + c $$ is as follows:
- The sum of the zeroes ($\alpha + \beta$) is given by $$ \frac{-b}{a} $$
- The product of the zeroes ($\alpha \beta$) is given by $$ \frac{c}{a} $$
Therefore, the correct answers are:
- A for the sum of the zeroes: $-b/a$
- B for the product of the zeroes: $c/a$
If $a$ and $b$ are two consecutive integers such that $a > b$ and the product of the two integers is $c$, then $(a + b)^{2} = $
(A) $1 - 4c$
B) $1 + 4c$ C) $1 - 2c$ (D) $1 + 2c$
Solution
Correct Option: $\mathbf{(B)} \ 1 + 4c$
Let $a$ and $b$ be two consecutive integers with $a > b$. This implies $a = b + 1$.
The product of $a$ and $b$ is given as $c$, hence: $$ c = a \times b = (b + 1) \times b = b^2 + b $$
The expression $(a + b)^2$ can be expanded and rewritten using the identities and given conditions: $$ (a + b)^2 = (b + 1 + b)^2 = (2b + 1)^2 $$ Expanding this expression, we get: $$ (2b + 1)^2 = 4b^2 + 4b + 1 $$
Given that $ab = c$, substituting $c = b^2 + b$ into the expanded expression, we find: $$ (2b + 1)^2 = 4(b^2 + b) + 1 = 4c + 1 $$
Therefore, $(a + b)^2 = \mathbf{1 + 4c}$, matching option $\mathbf{(B)}$.
If the line $y = mx + a$ meets the parabola $y^{2} = 4ax$ in two points whose abscissas are $x_{1}$ and $x_{2}$, then $x_{1} + x_{2} = 0$ if
(A) $m = -1$ B) $m = 1$ C) $m = 2$ D) $m = \frac{-1}{2}$
The correct answer is C) $m = 2$. Here is a step-by-step explanation of why this is the case:
Given the quadratic equation: $$ y^2 = 4ax $$ and substituting $y = mx + a$ into this equation, we have: $$ (mx + a)^2 = 4ax $$ Expanding and rearranging terms gives: $$ m^2x^2 + 2amx + a^2 - 4ax = 0 $$ Simplifying further: $$ m^2x^2 + (2am - 4a)x + a^2 = 0 $$ Sum of the roots of the quadratic equation $Ax^2 + Bx + C = 0$ is given by: $$ x_1 + x_2 = -\frac{B}{A} $$ Substituting $A = m^2$, $B = 2am - 4a$ into this formula: $$ x_1 + x_2 = -\frac{2am - 4a}{m^2} $$ To satisfy the condition $x_1 + x_2 = 0$, we solve: $$ -\frac{2am - 4a}{m^2} = 0 $$ This implies: $$ 2am - 4a = 0 $$ Factoring out $2a$ leads to: $$ 2a(m - 2) = 0 $$ Assuming $a \neq 0$, we find: $$ m - 2 = 0 \Rightarrow m = 2 $$ Therefore, the choice C) $m = 2$ ensures that the abscissas of the points where the line meets the parabola sum to zero.
The number of integral values of 'a' for which the quadratic equation $(x+a)(x+1991)+1=0$ has integral roots is:
A) 3
B) 0
C) 1
D) 2
Solution:
The correct option is D
To find the number of integral values of $a$ for which the quadratic equation $$(x+a)(x+1991)+1=0$$ has integral roots, we start by simplifying the equation:
$$ (x + a)(x + 1991) + 1 = 0 $$
This implies:
$$ (x + a)(x + 1991) = -1 $$
Now, considering that the product of two numbers is $-1$, there are specific pairs that fulfill this: Either one of the numbers is $1$ and the other is $-1$, or vice versa. Thus, we analyze these scenarios:
-
First Case: $x + a = 1$ and $x + 1991 = -1$
From these equations, by eliminating $x$, we get:
$$ a = 1 - 1991 - x = x - 1990 \Rightarrow x = -1991, \quad a = -1990 - 1991 = -3982 $$
But this mismatch shows an error in assumptions, implying this doesn't lead to a valid integral value of $a$.
-
Second Case: $x + a = -1$ and $x + 1991 = 1$
Solving these yields:
$$ a = -1 - x, \quad x = 1 - 1991 = -1990 $$
Hence:
$$ a = -1 - (-1990) = 1989 $$
-
Third Case: Switching the assignments from the second case:
$$ x + a = 1, \quad x + 1991 = -1 $$
Solving these gives:
$$ a = 1 - x, \quad x = -1 - 1991 = -1992 $$
Thus:
$$ a = 1 - (-1992) = 1993 $$
Conclusion: The values of $a$ that satisfy the equation are $1989$ and $1993$. Therefore, there are 2 integral values of $a$ that make the roots of the quadratic equation integral.
Find the roots of the quadratic equation $a x^{2}+b x+c=0$.
A) $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2a}$
B) $x=\frac{-b+\sqrt{b^{2}-4 a c}}{4a}$
C) $x=\frac{-b-\sqrt{b^{2}-4 a c}}{4a}$
D) $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2a}$
Solution The correct option is D) $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
The formula for the roots of a quadratic equation $ax^2 + bx + c = 0$ is presented as: $$ x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a} $$ This formula yields two possible roots: $\frac{-b+\sqrt{b^{2}-4ac}}{2a}$ and $\frac{-b-\sqrt{b^{2}-4ac}}{2a}$, which correspond to the plus and minus variations in the solution, respectively.
If the line $y=x$ is a tangent to the parabola $y=ax^{2}+bx+c$ at the point $(1,1)$ and the curve passes through $(-1,0)$, then
A) $a=b=-1$, $c=3$
B) $a=b=\frac{1}{2}$, $c=0$
C) $a=c=\frac{1}{4}$, $b=\frac{1}{2}$
D) $a=0$, $b=c=\frac{1}{2}$
To solve the problem, we start by analyzing the given conditions for the parabola $y = ax^2 + bx + c$ :
-
Line y = x is tangent to the parabola at (1,1):
Being a tangent line at $(1, 1)$, the slope of the tangent (which is the derivative of the parabola at this point) should equal the slope of the line $y = x$. Thus, we calculate the first derivative of $y$: $$ \frac{dy}{dx} = 2ax + b $$ At $(1,1)$, this becomes: $$ 1 = 2a \cdot 1 + b \quad \text{(equation 1)} $$
-
The parabola passes through (1,1):
Plugging this point into the equation of the parabola yields: $$ 1 = a(1)^2 + b(1) + c \quad \text{(equation 2)} $$
-
The parabola also passes through (-1,0):
Using this point in the parabola's equation gives: $$ 0 = a(-1)^2 - b(1) + c \quad \text{(equation 3)} $$
Now, solving equations 1, 2, and 3 together:
From equation 1: $$ b = 1 - 2a $$
Substitute $b$ in equation 2: $$ 1 = a + (1 - 2a) + c $$ $$ 1 = 1 - a + c \quad \Rightarrow \quad c = a $$
Using $c = a$ in equation 3: $$ 0 = a - (1 - 2a) + a $$ $$ 0 = 4a - 1 \quad \Rightarrow \quad a = \frac{1}{4} $$
With $a = \frac{1}{4}$, finding $b$: $$ b = 1 - 2 \times \frac{1}{4} = \frac{1}{2} $$
And since $c = a$, we have $c = \frac{1}{4}$.
Thus, we find the values:
- $a = \frac{1}{4}$
- $b = \frac{1}{2}$
- $c = \frac{1}{4}$
Option C ($a = c = \frac{1}{4}, b = \frac{1}{2}$) is correct.
What must be added to $\left(x^{2} + xy - y^{2}\right)$ to get $xy$?
A $y^{2}x^{2}$
B $y^{2} + x^{2}$
C $-y^{2} + x^{2}$
D $y^{2} - x^{2}$
To solve the question, "What must be added to $\left(x^2 + xy - y^2\right)$ to get $xy$?", we need to determine the value that, when added to $\left(x^2 + xy - y^2\right)$, results in $xy$.
Let's denote the expression to be added as $E$. We have: $$ x^2 + xy - y^2 + E = xy $$ To isolate $E$, rearrange this equation: $$ E = xy - (x^2 + xy - y^2) $$ Simplify the right-hand side: $$ \begin{align*} E &= xy - x^2 - xy + y^2 \ &= y^2 - x^2 \end{align*} $$
Thus, the needed expression to add is $y^2 - x^2$.
The correct answer is Option D: $y^2 - x^2$.
If both the roots of the quadratic equation $x^{2} - 2kx + k^{2} + k - 5 = 0$ are less than 5, then $k$ lies in the interval
A) $(5, 6]$
B) $(6, \infty)$
C) $(-\infty, 4)$
D) $[4, 5]$
The correct answer is C) $(-\infty, 4)$.
For the quadratic equation $$ f(x) = x^2 - 2kx + k^2 + k - 5, $$ both roots are less than 5. To ensure this, the conditions to be satisfied include:
- Discriminant $\geq 0$ (ensures real roots),
- $f(5) > 0$ (ensures $x = 5$ isn't a root and roots are less than 5),
- Average of roots $< 5$ (roots should be on the left of 5).
Let's evaluate each:
1. Discriminant $\geq 0$
For a quadratic $ax^2 + bx + c = 0$, the discriminant $\Delta = b^2 - 4ac$. Thus, $$ \Delta = (2k)^2 - 4 \cdot 1 \cdot (k^2 + k - 5) = 4k^2 - 4(k^2 + k - 5) = 4k^2 - 4k^2 - 4k + 20 \geq 0. $$ Simplifying, $$ -4k + 20 \geq 0 \Rightarrow k \leq 5. $$
2. $f(5) > 0$
Substituting $x = 5$ into the equation, $$ f(5) = 25 - 10k + k^2 + k - 5 = k^2 - 9k + 20 > 0. $$ Factoring, $$ (k-5)(k-4) > 0. $$ The inequality holds for $k \in (-\infty, 4) \cup (5, \infty)$.
3. Average of roots $< 5$
The sum of the roots for $ax^2 + bx + c = 0$ is given by $-\frac{b}{a}$. Therefore, the average of the roots is, $$ \frac{-(-2k)}{2} = k < 5. $$
Intersection of Conditions
Conditions from steps 1 and 3 constrain $k \leq 5$. However, from $f(5) > 0$, solutions must either be greater than 5 or less than 4. Combining all:
$$ k \in (-\infty, 4) \cap (-\infty, 5) = (-\infty, 4). $$
Thus, the suitable interval for $k$ is $(-\infty, 4)$.
Two parabolas with a common vertex and with axes along the $x$-axis and $y$-axis, respectively, intersect each other in the first quadrant. If the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is:
A) $4(x + y) + 3 = 0
B) $3(x + y) + 4 = 0
C) $8(2x + y) + 3 = 0
D) $x + 2y + 3 = 0
The correct answer is Option A: $4(x+y)+3=0$.
First, let's identify the equations of the parabolas. Given that both have a latus rectum of 3, and one has its axis along the $x$-axis while the other along the $y$-axis, their equations are: $$ x^2 = 3y \quad \text{and} \quad y^2 = 3x $$
We use a common tangent approach for solving this. The general tangent equation to the parabola $y^2 = 3x$ in the slope form is: $$ y = mx + \frac{3}{4m} $$
To ensure this line is also tangent to the parabola $x^2 = 3y$, substitute $y = mx + \frac{3}{4m}$ into $x^2 = 3y$ to get: $$ x^2 = 3\left(mx + \frac{3}{4m}\right) $$ Simplifying this leads to a quadratic in $x$: $$ x^2 - 3mx - \frac{9}{4m} = 0 $$
For this to be tangent, the discriminant (D) of this quadratic equation must be zero (indicating a single point of contact): $$ D = 0 \implies (3m)^2 - 4 \cdot 1 \cdot \left(-\frac{9}{4m}\right) = 9m^2 + \frac{36}{4m} = 0 $$ $$ m^3 + 1 = 0 \implies m = -1 \quad (\text{since } m \neq 0) $$
With $m = -1$, substituting back into the tangent formula gives: $$ y = -x - \frac{3}{4 \cdot (-1)} \implies y = -x + \frac{3}{4} $$ Expressed in standard linear form: $$ 4x + 4y + 3 = 0 \implies 4(x + y) + 3 = 0 $$
Thus, the equation of the common tangent to both parabolas is $4(x+y) + 3 = 0$.
The value of $b^2 - 4ac$ for the equation $3x^2 - 7x - 2=0$ is
A) 73
B) 25
C) 49
D) 0
The correct answer is A) 73.
Given equation: $$ 3x^2 - 7x - 2 = 0 $$ where (a = 3), (b = -7), and (c = -2).
Formula Used:
To find the value of $b^2 - 4ac$, we use the discriminant formula in quadratic equations.
Calculate: $$ b^2 - 4ac = (-7)^2 - 4 \cdot 3 \cdot (-2) $$ $$ = 49 + 24 $$ $$ = 73 $$
Thus, the value of $b^2 - 4ac$ is 73.
Let $f(x) = (\lambda^{2} + \lambda - 2) x^{2} + (\lambda + 2) x$ be a quadratic polynomial. The sum of all integral values of $\lambda$ for which $f(x) < 1$ for all $x \in \mathbb{R}$ is:
A) -1 B) -3 C) 0 D) -2
To determine the integral values of $\lambda$ such that the quadratic function $f(x) = (\lambda^2 + \lambda - 2) x^2 + (\lambda + 2) x < 1$ for all $x \in \mathbb{R}$, we start by rewriting the inequality in terms of another function $g(x)$:
$$ g(x) = (\lambda^2 + \lambda - 2) x^2 + (\lambda + 2) x - 1 < 0 $$
For $g(x)$ to be negative for all $x$, two main conditions must be satisfied:
- Leading coefficient $<0$
- Discriminant (D) $<0$
Condition 1: Leading Coefficient is Negative
We have the leading coefficient as $(\lambda^2 + \lambda - 2)$. This needs to be negative, implying:
$$ \lambda^2 + \lambda - 2 < 0 \ (\lambda + 2)(\lambda -1) < 0 \ \lambda \in (-2, 1) \quad \text{(1)} $$
Condition 2: Discriminant is Negative
The discriminant of $g(x)$, from $ax^2 + bx + c < 0$ being negative, where $a = \lambda^2 + \lambda - 2$ and $b = \lambda + 2$, gives:
$$ D = (\lambda + 2)^2 - 4(\lambda^2 + \lambda - 2)(-1) < 0 \ (\lambda + 2)^2 + 4(\lambda^2 + \lambda - 2) < 0 \ (\lambda + 2)(\lambda + 2 + 4\lambda - 4) < 0 \ (5\lambda - 2)(\lambda + 2) < 0 \ \lambda \in \left(-2, \frac{2}{5}\right) \quad \text{(2)} $$
Combining the conditions from (1) and (2):
$$ \lambda \in \left(-2, \frac{2}{5}\right) $$
We find the integral values of $\lambda$ within this range, which are $-1$ and $0$. Thus, the sum of these integral values is:
$$ -1 + 0 = -1 $$
Therefore, the sum of all integral values of $\lambda$ for which $g(x) < 0$ is -1. The correct answer is (A) -1.
If 2 is a root of the quadratic equation $3x^{2} + px - 8 = 0$ and the quadratic equation $4x^{2} - 2px + k = 0$ has equal roots, find the value of $k$.
Since 2 is a root of the quadratic equation $$3x^2 + px - 8 = 0,$$ we substitute $x = 2$ into the equation to get: $$ 3(2)^2 + p(2) - 8 = 0. $$
This simplifies to: $$ 12 + 2p - 8 = 0. $$
Further simplification gives: $$ 4 + 2p = 0 \ 2p = -4 \ p = -2. $$
With $p = -2$, the second quadratic equation $$4x^2 - 2px + k = 0$$ becomes: $$ 4x^2 + 4x + k = 0. $$
For a quadratic equation to have equal roots, the discriminant (denoted as $D$) must be zero. The discriminant is calculated as: $$ D = b^2 - 4ac. $$
For the equation $4x^2 + 4x + k = 0$, the coefficients are $a = 4$, $b = 4$, and $c = k$. Substituting these into the discriminant formula gives: $$ D = 4^2 - 4 \times 4 \times k = 16 - 16k = 0. $$
Solving for $k$, we get: $$ 16 = 16k \ k = 1. $$
Thus, the value of $k$ such that the quadratic equation $4x^2 + 4x + k = 0$ has equal roots is $k = 1$.
Let $S$ be the set of values of 'a' for which 2 lies between the roots of the quadratic equation $x^{2} + (a + 2)x - (a + 3) = 0$. Then $S$ is given by:
A $(-\infty, -5)$
B $(5, \infty)$
C $(-\infty, -5]$
D $[5, \infty)$
Solution
To determine the set $S$ for which 2 lies between the roots of the quadratic equation given, we need to check when the equation $$x^2 + (a+2)x - (a+3) = 0$$ has 2 as an interior point between its roots.
Hence, calculate the value of the quadratic equation at $x = 2$: $$ f(2) = 2^2 + (a+2)\cdot2 - (a+3) $$ Simplify this: $$ f(2) = 4 + 2a + 4 - a - 3 = a + 5 $$ For 2 to lie between the roots, the value of $f(2)$ must be less than 0: $$ a + 5 < 0 $$ Simplify: $$ a < -5 $$ This inequality describes values of $a$ such that 2 is between the roots. Therefore, the set $S$ is: $$ \boldsymbol{S = (-\infty, -5)} $$ Thus, the correct answer is Option A ($(-\infty, -5)$).
If $\alpha, \beta$ are the roots of the equation $3x^2 + 5x + 4 = 0$, then the quadratic equation whose roots are $\alpha^{2}, \beta^{2}$
(A) $9x^2 + x + 16 = 0$
(B) $9x^2 + x + 24 = 0$
(C) $9x^2 - x + 16 = 0$
(D) $9x^2 - x + 24 = 0$
The correct answer is (C) $9x^2 - x + 16 = 0$.
Let's derive that starting with the original quadratic equation $3x^2 + 5x + 4 = 0$. The sum $(\alpha + \beta)$ and product $(\alpha \beta)$ of the roots can be calculated using Vieta's formulas: $$ \alpha+\beta = -\frac{b}{a} = -\frac{5}{3}, \quad \alpha\beta = \frac{c}{a} = \frac{4}{3} $$
Objective: Find the quadratic equation with roots $\alpha^2$ and $\beta^2$.
From basic algebraic identities: $$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(-\frac{5}{3}\right)^2 - 2\left(\frac{4}{3}\right) $$ Calculating the above, we find: $$ \alpha^2 + \beta^2 = \frac{25}{9} - \frac{8}{3} = \frac{25 - 24}{9} = \frac{1}{9} $$
Next, the product $\alpha^2 \beta^2$: $$ \alpha^2 \beta^2 = (\alpha \beta)^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} $$
The derived quadratic equation having roots $\alpha^2$ and $\beta^2$ will be: $$ x^2 - (\alpha^2+\beta^2)x + \alpha^2\beta^2 = 0 $$
Substituting the values, we get: $$ x^2 - \frac{1}{9}x + \frac{16}{9} = 0 $$
Multiplying through by 9 to clear the fractions: $$ 9x^2 - x + 16 = 0 $$
Thus, the quadratic equation whose roots are $\alpha^2$ and $\beta^2$ is $9x^2 - x + 16 = 0$, matching option (C).
If $x^{2}=k y^{2}$, where $x$ and $y$ are whole numbers, then which of the following is always true?
A. $x > y$
B. $x < y$
C. $\mathrm{k}$ is an integer
D. $\mathrm{k}$ is a non-negative number
Solution
The correct option is D.
Given the equation $$x^2 = ky^2,$$ where $x$ and $y$ are whole numbers, we can imply several things:
- Since $x^2$ and $y^2$ are squares of whole numbers, they are always non-negative.
- Therefore, $k$, computed as $$k = \frac{x^2}{y^2},$$ must also be non-negative since both the numerator ($x^2$) and the denominator ($y^2$) are non-negative.
This leads us to conclude that option D ($k$ is a non-negative number) is always true, regarding the values of $x$ and $y$. Options A and B about the size comparison between $x$ and $y$ depend on the value of $k$, and C (about $k$ being an integer) is not necessarily true without additional information about the relationship between $x$ and $y$ beyond their being whole numbers.
If $r, s$ are the roots of the equation $x^{2}-px+q=0$, then what is the equation of the curve whose roots are $\left(\left[\frac{1}{r}\right]+s\right)$ and $\left(\left[\frac{1}{5}\right]+r\right)$?
(A) $\mathrm{qx}^{2}+(\mathrm{p}-\mathrm{q})x+(\mathrm{q}^{2}-1)=0$
(B) $(p-q)x^{2}+pqx-(p+1)(q+1)=0$
(C) $qx^{2}-p(q+1)x+(1+q)^{2}=0$
(D) $qx^{2}-p(q+1)x+(1+q)=0$
The correct option is (C) $qx^2 - p(q+1)x + (1+q)^2 = 0$.
To verify this, we can use a particular set of values for $r$ and $s$:
- Let $r = 1$ and $s = 1$, then $p = r + s = 2$ and $q = rs = 1$.
The roots of the curve according to the original equation would then be:
- $\left(\left[\frac{1}{r}\right]+s\right) = ([1] + 1) = 2$
- $\left(\left[\frac{1}{s}\right]+r\right) = ([1] + 1) = 2$
The equation with roots $2, 2$ must be: $$ (x-2)^2 = 0 \implies x^2 - 4x + 4 = 0 $$
Substitute $p = 2$ and $q = 1$ into each option to determine which reproduces this confirmed result:
- Option (C): $qx^2 - p(q+1)x + (1+q)^2 = 0$ $$ 1 \cdot x^2 - 2(1+1) \cdot x + (1+1)^2 = 0 \implies x^2 - 4x + 4 = 0 $$
This aligns perfectly, confirming option (C) as the correct answer.
If $a, \beta$ are the roots of $a x^{2}+2 b x+c=0$ and $a+\delta, \beta+\delta$ are those of $A x^{2}+2 B x+C=0$, then the value of $\frac{\left(b^{2}-a c\right)}{\left(B^{2}-A C\right)}$ is
A $\left(\frac{a}{A}\right)^{2}$
B $\left(\frac{a}{a}\right)^{2}$
C 1
The correct option is A $\left(\frac{a}{A}\right)^{2}$.
Consider the roots of the given quadratic equations. If the roots of the second equation are denoted as $\alpha', \beta'$, then they are related to the roots of the first equation by: $$ \alpha' = a + \delta, \quad \beta' = \beta + \delta $$ Thus, the differences of the roots remain the same: $$ \alpha' - \beta' = (a + \delta) - (\beta + \delta) = a - \beta $$ Furthermore, the sum and product of roots of the equations can be analyzed through the quadratic relationships: $$ \alpha + \beta = -\frac{2b}{a}, \quad \alpha \beta = \frac{c}{a} $$ $$ \alpha' + \beta' = -\frac{2B}{A}, \quad \alpha' \beta' = \frac{C}{A} $$ These lead us to establish the relation between the discriminants of these quadratic equations: $$ (\alpha + \beta)^2 - 4\alpha\beta = \frac{(2b)^2 - 4ac}{a^2} = \frac{4(b^2-ac)}{a^2} $$ and $$ (\alpha' + \beta')^2 - 4\alpha'\beta' = \frac{(2B)^2 - 4AC}{A^2} = \frac{4(B^2-AC)}{A^2} $$ Equating the normalized discriminants: $$ \frac{4(b^2-ac)}{a^2} = \frac{4(B^2-AC)}{A^2} $$
Simplifying, we get: $$ \frac{b^2-ac}{B^2-AC} = \frac{a^2}{A^2} $$
Thus, the desired ratio $\frac{b^2-ac}{B^2-AC}$ is: $$ \left(\frac{a}{A}\right)^2 $$
Hence, the correct answer is A $\left(\frac{a}{A}\right)^{2}$.
If $f: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $f(x) = x^{2} - 3x + 2$, find $f(f(x))$.
Solution
Given the function $f:\mathbb{R} \rightarrow \mathbb{R}$ defined by: $$ f(x) = x^2 - 3x + 2. $$ We need to find $f(f(x))$. To do this, first, we substitute $f(x)$ into $f$ again:
$$ f(f(x)) = f(x^2 - 3x + 2). $$
Substitute $x^2 - 3x + 2$ into $f$'s original formula: $$ f(x^2 - 3x + 2) = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2. $$
First, expand $(x^2 - 3x + 2)^2$: $$ (x^2 - 3x + 2)^2 = x^4 - 6x^3 + 9x^2 + 4x^2 -12x + 4. $$
Combine like terms: $$ = x^4 - 6x^3 + 13x^2 -12x + 4. $$
Next, expand $-3(x^2 - 3x + 2)$: $$ -3(x^2 - 3x + 2) = -3x^2 + 9x - 6. $$
Combine all components: $$ x^4 - 6x^3 + 13x^2 - 12x + 4 - 3x^2 + 9x - 6 + 2. $$
Simplify by combining like terms: $$ x^4 - 6x^3 + 10x^2 - 3x + 0. $$
Therefore, the required function composition $f(f(x))$ simplifies to: $$ f(f(x)) = x^4 - 6x^3 + 10x^2 - 3x. $$
Conclusion: $$ f(f(x)) = x^4 - 6x^3 + 10x^2 - 3x. $$
For a quadratic equation of the form $a x^{2} + b x + c = 0$, the solutions are $x_{1,2} = \frac{-b \pm \sqrt{b^{2} - 4 a c}}{2 a}$.
To solve the quadratic equation in the form $$ a x^{2} + b x + c = 0 $$ given the coefficients $a=1$, $b=10$, and $c=-25$, we apply the quadratic formula $$ x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$.
Substituting the values of $a$, $b$, and $c$ into the formula, we have:
$$ x_{1,2} = \frac{-10 \pm \sqrt{10^2 - 4 \times 1 \times (-25)}}{2 \times 1} $$
This simplifies to:
$$ x_{1,2} = \frac{-10 \pm \sqrt{100 + 100}}{2} $$
$$ x_{1,2} = \frac{-10 \pm \sqrt{200}}{2} $$
Since $\sqrt{200} = 10\sqrt{2}$, the expression further simplifies to:
$$ x_{1,2} = \frac{-10 \pm 10\sqrt{2}}{2} $$
Breaking this into two solutions:
- For the positive square root: $$ x_1 = \frac{-10 + 10\sqrt{2}}{2} = 5(-1 + \sqrt{2}) = 5(1 + \sqrt{2}) $$
- For the negative square root: $$ x_2 = \frac{-10 - 10\sqrt{2}}{2} = 5(-1 - \sqrt{2}) = 5(1 - \sqrt{2}) $$
The final solutions to the quadratic equation are: $$ x = 5(1 + \sqrt{2}), , x = 5(1 - \sqrt{2}) $$
$6x^2 - x - 2 = 0, x = -1/2, 2/3$
To solve the quadratic equation $$6x^2 - x - 2 = 0,$$ we start by factoring: Firstly, we rewrite the quadratic as: $$ 6x^2 + 3x - 4x - 2 = 0. $$ Here, we split the middle term ($-x$) into $3x - 4x$ to facilitate factoring by grouping. Next, group the terms: $$ 3x(2x + 1) - 2(2x + 1) = 0. $$ Notice that $(2x + 1)$ is a common factor. Therefore, factor it out: $$ (3x - 2)(2x + 1) = 0. $$ We have a product of two factors equaling zero. According to the zero-product property, one or both of the factors must be zero. So, we get: $$ 3x - 2 = 0 \quad \text{or} \quad 2x + 1 = 0. $$ Solving these equations individually:
- For $3x - 2 = 0$: $$ 3x = 2 \implies x = \frac{2}{3}. $$
- For $2x + 1 = 0$: $$ 2x = -1 \implies x = -\frac{1}{2}. $$
Hence, the solutions to the equation $6x^2 - x - 2 = 0$ are: $$x = \frac{2}{3} \quad \text{and} \quad x = -\frac{1}{2}.$$
For what value of $p$, $x=3$ will be a root of the equation $p x^{2} + 2x - 3 = 0$?
(A) $1/3$
B) 3
C) -3
(D) $-1/3$
The correct value of ( p ) for which ( x = 3 ) is a root of the equation ( p x^{2} + 2x - 3 = 0 ) is given by option (D) $-\frac{1}{3}$.
To find this, substitute ( x = 3 ) into the equation: $$ p \cdot 3^2 + 2 \cdot 3 - 3 = 0 $$ This simplifies to: $$ 9p + 6 - 3 = 0 $$ Further reduction gives: $$ 9p + 3 = 0 $$ Solving for ( p ), we isolate ( p ) on one side: $$ 9p = -3 $$ Thus, ( p ) evaluates to: $$ p = -\frac{1}{3} $$ This confirms that the value of ( p ) that makes ( x = 3 ) a root of the given quadratic equation is ( \boxed{-\frac{1}{3}} ).
For what values of $p$ are the roots of the equation $4x^{2} + px + 3 = 0$ real and equal?
For the equation $$ 4x^2 + px + 3 = 0 $$ to have real and equal roots, the discriminant must be zero. The discriminant ($\Delta$) for a quadratic equation $ax^2 + bx + c = 0$ is given by: $$ \Delta = b^2 - 4ac. $$
For our equation, we have $a = 4$, $b = p$, and $c = 3$. Plugging these values into the discriminant formula gives: $$ \Delta = p^2 - 4 \times 4 \times 3 = p^2 - 48. $$
Setting $\Delta$ to zero for real and equal roots, we have: $$ p^2 - 48 = 0, $$ which simplifies to: $$ p^2 = 48. $$
Therefore, solving for $p$, we take the square root: $$ p = \pm \sqrt{48}. $$
We can simplify $\sqrt{48}$ as: $$ p = \pm \sqrt{16 \times 3} = \pm 4 \sqrt{3}. $$
Thus, the values of $p$ for which the roots are real and equal are $p = \pm 4 \sqrt{3}$.
If the curve $y = ax^{2} + bx + c$, $x \in \mathbb{R}$, passes through the point $(1, 2)$ and the tangent line to this curve at the origin is $y = x$, then the possible values of $a, b, c$ are
A) $a = 1$, $b = 1$, $c = 0$
B) $a = -1$, $b = 1$, $c = 1$
C) $a = 1$, $b = 0$, $c = 1$
D) $a = \frac{1}{2}$, $b = \frac{1}{2}$, $c = 1$
To solve the problem, observe that:
-
The curve $$y = ax^2 + bx + c$$ must pass through the point $(1, 2)$. This condition yields the equation: $$ 2 = a \cdot 1^2 + b \cdot 1 + c \Rightarrow a + b + c = 2. $$
-
The tangent to the curve at the origin $(0, 0)$ is given as $y = x$. The condition that the tangent line at the origin is $y = x$ implies that the derivative of the curve at $x = 0$ should equal $1$. This leads to: $$ \frac{dy}{dx} = 2ax + b \Rightarrow \left. \frac{dy}{dx} \right |_{x=0} = b. $$ Since we know the tangent at $x = 0$ is $y = x$, we find that $b = 1$.
-
Substituting $b = 1$ into $a+b+c = 2$, we get: $$ a + 1 + c = 2 \Rightarrow a + c = 1. $$
-
To check additional points, use the fact that the curve passes through the origin $(0,0)$: $$ 0 = a \cdot 0^2 + 1 \cdot 0 + c \Rightarrow c = 0. $$
-
Substituting $c = 0$ into $a + c = 1$, we get: $$ a + 0 = 1 \Rightarrow a = 1. $$
Thus, $a = 1$, $b = 1$, and $c = 0$ are the values that satisfy all given conditions.
Selecting the correct option from the choices:
- A) $a = 1$, $b = 1$, $c = 0$ (This matches our calculated values)
Option A is correct: $a = 1$, $b = 1$, $c = 0$. This solution is confirmed by analyzing both the conditions given for the point the curve passes through and the slope of the tangent at the origin.
Let $p, q \in {1, 2, 3, 4}$. The number of equations of the form $p x^{2} + q x + 1 = 0$ having real roots is
(A) 15
(B) 9
(C) 7
(D) 8
Solution
The correct answer is (C) 7.
In order for the quadratic equation $p x^2 + q x + 1 = 0$ to have real roots, the discriminant must be non-negative:
$$ \boldsymbol{q^2 - 4p \geq 0} \Rightarrow q^2 \geq 4p $$
Considering each value of $p$:
- For $\boldsymbol{p = 1}$, we need $\boldsymbol{q^2 \geq 4}$. Possible values for $q$ are $\boldsymbol{2, 3, 4}$.
- For $\boldsymbol{p = 2}$, we need $\boldsymbol{q^2 \geq 8}$. Possible values for $q$ are $\boldsymbol{3, 4}$.
- For $\boldsymbol{p = 3}$, we need $\boldsymbol{q^2 \geq 12}$. The only possible value for $q$ is $\boldsymbol{4}$.
- For $\boldsymbol{p = 4}$, we need $\boldsymbol{q^2 \geq 16}$. The only possible value for $q$ is $\boldsymbol{4}$.
Counting all valid combinations, we find a total of $3$ solutions for $p=1$, $2$ solutions for $p=2$, $1$ solution for $p=3$, and $1$ solution for $p=4$, leading to:
$$ \textbf{Total of 7 valid solutions} $$
A quadratic equation whose roots are $2+\frac{1}{\sqrt{2}}$ and $2-\frac{1}{\sqrt{2}}$ is:
A) $x^{2}-7x+8=0$ B) $2x^{2}-8x+7=0$ C) $2x^{2}-7x+8=0$ D) $x^{2}-8x+7=0
To find the quadratic equation with given roots, we use the relationships between the roots and coefficients of the equation given by Vieta's formulas. For a quadratic equation of the form $x^2 + bx + c = 0$, the sum of the roots ($\alpha + \beta$) and the product of the roots ($\alpha \beta$) are:
- Sum of the roots $\alpha + \beta = -b$
- Product of the roots $\alpha \beta = c$
Given Roots: $$ \alpha = 2 + \frac{1}{\sqrt{2}}, \quad \beta = 2 - \frac{1}{\sqrt{2}} $$
Calculate the sum and product of the roots:
- Sum of the roots: $$ \alpha + \beta = \left(2 + \frac{1}{\sqrt{2}}\right) + \left(2 - \frac{1}{\sqrt{2}}\right) = 4 $$
- Product of the roots: $$ \alpha \beta = \left(2 + \frac{1}{\sqrt{2}}\right)\left(2 - \frac{1}{\sqrt{2}}\right) $$ Applying the difference of squares formula, $a^{2} - b^{2} = (a+b)(a-b)$: $$ \alpha \beta = 4 - \left(\frac{1}{\sqrt{2}}\right)^2 = 4 - \frac{1}{2} = \frac{7}{2} $$
The quadratic equation is thus formed as follows: $$ x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 $$ $$ x^2 - 4x + \frac{7}{2} = 0 $$
To convert this equation so that all terms have integer coefficients, multiply through by 2: $$ 2x^2 - 8x + 7 = 0 $$
Thus, option B is correct: $$ \textbf{B) } 2x^{2}-8x+7=0 $$
Condition for $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0$ to be a quadratic equation is
A $a \geq 0; a, b, c$ are real
B $a < 0; a, b, c$ are unreal
C $\mathrm{a} \leq \mathrm{0}; \mathrm{a}, \mathrm{b}, \mathrm{c}$ are real
D $a \neq 0; a, b, c$ are real.
Solution
The correct choice is Option D: $a \neq 0; a, b, c$ are real.
For an equation to be classified as a quadratic equation, the following conditions must be met:
- The equation must be of the second degree, specifically in the variable $x$. This is characterized by the highest power of $x$ being 2, which means the coefficient of $x^2$, denoted as $a$, must be non-zero (i.e., $a \neq 0$).
- The coefficients $a, b, c$ must all be real numbers to satisfy the typical definition of a quadratic equation in real numbers.
Thus, the conditions $a \neq 0$ and that $a, b$, and $c$ are real are necessary to ensure the equation $ax^2 + bx + c = 0$ is a valid quadratic equation.
If $x=\frac{2}{3}$ is a solution of the quadratic equation $7x^{2}+mx-3=0$, find the value of $m$.
(A) $m=\frac{1}{6}$
(B) $m=\frac{-1}{3}$
(C) $m=\frac{-1}{6}$
(D) $m=\frac{1}{3}$
Given: If ( x = \frac{2}{3} ) is a solution of the quadratic equation ( 7x^2 + mx - 3 = 0 ).
To find ( m ), substitute ( x = \frac{2}{3} ) into the equation:
-
Substitute and simplify: $$ 7\left(\frac{2}{3}\right)^2 + m\left(\frac{2}{3}\right) - 3 = 0 $$ $$ 7\left(\frac{4}{9}\right) + \frac{2m}{3} - 3 = 0 $$ $$ \frac{28}{9} + \frac{2m}{3} - 3 = 0 $$
-
Convert all terms to a common denominator and simplify: $$ \frac{28}{9} + \frac{6m}{9} - \frac{27}{9} = 0 $$ $$ 28 + 6m - 27 = 0 $$
-
Solve for ( m ): $$ 6m + 1 = 0 $$ $$ 6m = -1 $$ $$ m = \frac{-1}{6} $$
Therefore, ( m = \frac{-1}{6} ), and the correct answer is (C).
The function $f(x) = x^{2}(x-2)^{2}$.
A. Decreases on $(0, 1) \cup (2, \infty)$.
B. Increases on $(-\infty, 0) \cup (1, 2)$.
C. Has a local maximum value of 0.
D. Has a local maximum value of 1.
The correct answer is D: Has a local maximum value of 1.
Firstly, consider the given function: $$ f(x) = x^2 (x - 2)^2 $$
We need to find the critical points by differentiating $f(x)$ and setting the derivative $f'(x)$ equal to zero. The derivative is calculated as follows: $$ f'(x) = 2x(x - 2)^2 + x^2(2(x - 2)) $$ Simplifying, we get: $$ f'(x) = 2x(x - 2)^2 + 2x^2(x - 2) = 4x(x - 1)(x - 2) $$
Setting $f'(x) = 0$, we solve for $x$: $$ 4x(x - 1)(x - 2) = 0 $$ The roots are $x = 0$, $x = 1$, and $x = 2$.
For determining whether these points are minima, maxima, or neither, we analyze the sign of $f'(x)$:
- For $x < 0$, $f'(x)$ is negative as all factors are negative.
- In the interval $(0, 1)$, $f'(x)$ is negative since two of the factors are positive and one is negative.
- At $x = 1$, $f'(x) = 0$ but changes sign from negative to positive on either side, suggesting a local minimum.
- For $x > 2$, $f'(x)$ is positive as all factors are positive.
Therefore, without detailed computation, error in the initial interpretation indicating a local maximum at $x = 1$ could imply confusing the root's behavioral analysis or computational error. On review, $x = 1$ is a local minimum since $f'(x)$ changes from negative to positive which contradicts answer D directly, and C. Has a local maximum value of 0 seems correct as $f(x)$ attains a maximum value of $0$ at $x = 0$ and $x = 2$.
Thus, typically, computation should reveal that option C is correct: the function has local maxima at $x = 0$ and $x = 2$, not $x = 1$. D appears incorrect unless there is more specific context or construal not provided.
For what value of $p$ does the quadratic equation $px^{2} - 18x + 1 = 0$ form a perfect square?
To find the value of $p$ for which the equation $px^2 - 18x + 1 = 0$ forms a perfect square, we have to ensure that the discriminant of the quadratic equation is zero. The discriminant for a quadratic equation $ax^2 + bx + c = 0$ is given by:
$$ \Delta = b^2 - 4ac $$
For the equation in question, $a = p$, $b = -18$, and $c = 1$. Plugging in these values into the discriminant formula yields:
$$ \Delta = (-18)^2 - 4 \cdot p \cdot 1 = 324 - 4p $$
Setting the discriminant to zero for the quadratic to be a perfect square, we solve for $p$:
$$ 324 - 4p = 0 $$
Solving for $p$, we get:
$$ 4p = 324 \ p = \frac{324}{4} \ p = 81 $$
Therefore, the value of $p$ that makes the quadratic equation $px^2 - 18x + 1 = 0$ a perfect square is $p = 81$. This results in the equation having double (repeated) roots, similar to the squared equation $(x + 2)^2 = 0$ which has a root of $x = -2$ repeated twice. Thus, the discriminant should indeed be zero.
If the three equations $x^{2}+ax+12=0$, $x^{2}+bx+15=0$, and $x^{2}+(a+b)x+36=0$ have a common possible root, then the sum of roots is:
A) 24
B) -24
C) 20
D) -20
The problem presents three quadratic equations and states that they have a common root. Let's denote the common root as $a$. The other roots of these equations are $\beta$, $\gamma$, and $\delta$, for each equation respectively. Using Vieta's formulas, we can write the following relationships based on the equations provided:
- For the equation $x^{2}+ax+12=0$: $$ a + \beta = -a \quad \text{and} \quad a \beta = 12 $$
- For the equation $x^{2}+bx+15=0$: $$ a + \gamma = -b \quad \text{and} \quad a \gamma = 15 $$
- For the equation $x^{2}+(a+b)x+36=0$: $$ a + \delta = -(a+b) \quad \text{and} \quad a \delta = 36 $$
Combining the sums from the first two equations, we get: $$ (a + \beta) + (a + \gamma) = -a - b $$ $$ = a + \delta \quad \text{(from the third equation)} $$ This results in: $$ a + \beta + \gamma = \delta $$
Multiplying and combining terms: $$ a(\beta + \gamma + \delta) = 12 + 15 + 36 = 63 $$ $$ \text{Using the equation } a + \beta + \gamma = \delta, \text{ we substitute } \delta: $$ $$ a(2\delta - a) = 63 $$ $$ 2a\delta - a^2 = 63 $$ Given $a\delta = 36$, we can substitute: $$ 72 - a^{2} = 63 $$ $$ a^2 = 9 $$ Thus, $a = \pm 3$.
For $a = 3$, we find the other roots as follows: $$ \beta = 4, \quad \gamma = 5, \quad \delta = 12 $$ Then, the sum of roots considering all equations: $$ a + \beta + \gamma + \delta = 24 $$
For $a = -3, similarly: $$ \beta = -4, \quad \gamma = -5, \quad \delta = -12 $$ And thus, the sum of these roots: $$ a + \beta + \gamma + \delta = -24 $$
Hence, the possible sums of the roots are:
A) 24 and B) -24
Consider the quadratic equation $(c-5) x^{2} - 2cx + (c-4) = 0$. Let $S$ be the set of all integral values of $c$ for which one root of the equation lies in the interval $(0,2)$ and another root lies in the interval $(2,3)$. The number of elements in $S$ is:
A. 18
B. 12
C. 11
D. 10
The correct answer is Option C: 11.
The given quadratic equation is: $$ (c-5) x^{2} - 2cx + (c-4) = 0 $$
First, we clarify that this formula should be quadratic, which requires: $$ c-5 \neq 0 \Rightarrow c \neq 5 $$
We can rewrite the original equation as: $$ x^2 - \frac{2c}{c-5}x + \frac{c-4}{c-5} = 0 $$
To satisfy the conditions that roots lie in intervals $(0, 2)$ and $(2, 3)$, we study the quadratic's value at specific points:
-
At $x=0$ and $x=2$: $$ f(0) \cdot f(2) < 0 \Rightarrow \frac{c-4}{c-5} \cdot \left(4 - \frac{4c}{c-5} + \frac{c-4}{c-5}\right) < 0 $$ Simplifying, this yields: $$ \frac{(c-4)(c-24)}{(c-5)^2} < 0 $$ Indicating that: $$ c \in (4, 24) - {5} $$
-
Between $x=2$ and $x=3$: $$ f(2) \cdot f(3) < 0 \Rightarrow \frac{(c-24)(4c-49)}{(c-5)^2} < 0 $$ This implies: $$ c \in \left(\frac{49}{4}, 24\right) $$
Combining these conditions, we find: $$ c \text{ must be in } \left(\frac{49}{4}, 24\right) \cap (4, 24) - {5} $$ This results in: $$ c \in {13, 14, \ldots, 23} $$
Thus, the number of potential values for $c$ is: $$ n(S) = 11 $$
This validates the choice, Option C: 11, as the number of integral solutions for $c$.
What is the sum of the roots of the quadratic equation $x^{2} - 2x - 15 = 0$?
The sum of the roots of the quadratic equation can be found using the formula derived from the standard form of a quadratic equation $ax^2 + bx + c = 0$. Given the equation: $$ x^2 - 2x - 15 = 0. $$
In this equation:
$a = 1$ (coefficient of $x^2$)
$b = -2$ (coefficient of $x$)
$c = -15$ (constant term)
According to Vieta's formulas, the sum of the roots of the quadratic equation is given by: $$ \text{sum of the roots} = -\frac{b}{a}. $$
Substituting the values of $b$ and $a$ from the equation, we get: $$ \text{sum of the roots} = -\frac{-2}{1} = 2. $$
Thus, the sum of the roots of the equation $x^2 - 2x - 15 = 0$ is 2.
What is the value of '$p$' for which the quadratic equation $2px^{2} + 6x + 5 = 0$ has equal roots?
To find the value of '$p$' for which the quadratic equation $2px^2 + 6x + 5 = 0$ has equal roots, we utilize the fact that for a quadratic equation of the form $$ax^2 + bx + c = 0,$$ the condition for equal roots is given by the discriminant, $D$, being equal to zero, where:
$$ D = b^2 - 4ac $$
In our equation, comparing terms:
$a = 2p$
$b = 6$
$c = 5$
Substitute these into the discriminant formula:
$$ D = (6)^2 - 4 \cdot (2p) \cdot 5 = 0 $$
Solve the equation:
$$ 36 - 40p = 0 $$
Rearrange to solve for $p$:
$$ 40p = 36 \ p = \frac{36}{40} $$
Simplify the fraction:
$$ p = \frac{9}{10} $$
Thus, the value of $p$ for which the quadratic equation has equal roots is $p = \frac{9}{10}$.
Find the zeros of $2x^2 - 5x + 3$.
To find the zeros of the quadratic equation
$$
2x^2 - 5x + 3 = 0
$$,
we can use the method of factorization.
Firstly, multiply the coefficient of $x^2$ (which is 2) and the constant term (which is 3), resulting in 6. Next, identify two factors of 6 that add up to the coefficient of the $x$ term (which is -5). The factors -2 and -3 satisfy this condition since $$ -2 + (-3) = -5 $$.
Now, write the middle term of the quadratic expression as the sum of -2x and -3x. The equation becomes: $$ 2x^2 - 2x - 3x + 3 = 0 $$.
Factor by grouping:
From the first two terms $2x^2 - 2x$, factor out $2x$, giving $2x(x - 1)$.
From the last two terms $-3x + 3$, factor out $-3$, giving $-3(x - 1)$.
Now, the equation looks like: $$ 2x(x - 1) - 3(x - 1) = 0 $$.
Factor out $(x - 1)$ from both groups: $$ (x - 1)(2x - 3) = 0 $$.
Set each factor equal to zero and solve for $x$:
$x - 1 = 0 \Rightarrow x = 1$
$2x - 3 = 0 \Rightarrow 2x = 3 \Rightarrow x = \frac{3}{2}$
Thus, the zeros of the equation $2x^2 - 5x + 3 = 0$ are $x = 1$ and $x = \frac{3}{2}$.
$\sqrt{2}x^{2} + 7x + 5\sqrt{2} = 0$
To solve the quadratic equation given:
$$ \sqrt{2}x^{2} + 7x + 5\sqrt{2} = 0 $$
We start by finding factors of the quadratic equation. It is noticeable that $\sqrt{2}$ is a common term in two of the terms, but we need a systematic approach involving the use of the quadratic formula. The quadratic formula is given by:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
In the equation $$\sqrt{2}x^{2} + 7x + 5\sqrt{2} = 0$$, the coefficients are $a = \sqrt{2}$, $b = 7$, and $c = 5\sqrt{2}$.
Plugging these values into the quadratic formula:
$$ x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot \sqrt{2} \cdot 5\sqrt{2}}}{2\sqrt{2}} $$
$$ x = \frac{-7 \pm \sqrt{49 - 40}}{2\sqrt{2}} $$
$$ x = \frac{-7 \pm \sqrt{9}}{2\sqrt{2}} $$
$$ x = \frac{-7 \pm 3}{2\sqrt{2}} $$
This results in two potential solutions for $x$:
$$x = \frac{-7 + 3}{2\sqrt{2}} = \frac{-4}{2\sqrt{2}} = \frac{-2}{\sqrt{2}} = -\sqrt{2}$$
$$x = \frac{-7 - 3}{2\sqrt{2}} = \frac{-10}{2\sqrt{2}} = \frac{-5}{\sqrt{2}} = -\frac{5\sqrt{2}}{2}$$
The solution for the given quadratic equation is: $$x = -\sqrt{2}, -\frac{5\sqrt{2}}{2}$$
Find a quadratic equation whose zeros are -7 and 5.
In this question, we are provided with the zeros of a polynomial, which are -7 and 5, and we need to find the quadratic equation based on these values.
First, let's determine the sum of the zeros: $$ \text{Sum of zeros} = -7 + 5 = -2 $$
Next, we calculate the product of the zeros: $$ \text{Product of zeros} = (-7) \times 5 = -35 $$
With these values, we can now construct the quadratic equation. The standard form of a quadratic equation based on its zeros can be given by: $$ y = x^2 - (\text{sum of zeros}) \times x + (\text{product of zeros}) $$ Substituting the values we calculated: $$ y = x^2 - (-2)x + (-35) $$ On simplifying, we get: $$ y = x^2 + 2x - 35 $$
This becomes the quadratic equation modelled from the zeros -7 and 5.
Solve for $x: $ $x^{4} - 20x^{2} + 64 = 0$.
To find the solution to the equation (x^4 - 20x^2 + 64 = 0), we can start by recognizing that it can be approached as a quadratic equation in terms of another variable ( y = x^2 ). Writing the equation in terms of ( y ):
$$ y^2 - 20y + 64 = 0. $$
Next, we'll factorize this quadratic equation. We search for two numbers that multiply to (64) (the constant term) and add up to (-20) (the coefficient of the (y) term). Those numbers are (-16) and (-4), since:
$$ -16 \times -4 = 64 \quad \text{and} \quad -16 + (-4) = -20. $$
Using these numbers, we can factorize our quadratic equation as follows:
$$ y^2 - 16y - 4y + 64 = 0 \ y(y - 16) - 4(y - 16) = 0 \ (y - 16)(y - 4) = 0. $$
Setting each factor to zero gives us the possible values for (y):
$$ y - 16 = 0 \quad \text{or} \quad y - 4 = 0 \ y = 16 \quad \text{or} \quad y = 4. $$
Recall that (y = x^2). Therefore, replacing (y) with (x^2), we have:
$$ x^2 = 16 \quad \text{or} \quad x^2 = 4. $$
Next, find (x) by taking the square root of both sides:
$$ x = \pm \sqrt{16} \quad \text{or} \quad x = \pm \sqrt{4} \ x = \pm 4 \quad \text{or} \quad x = \pm 2. $$
Thus, the values of (x) that satisfy the given equation are ( x = 4, -4, 2, -2 ).
If $\alpha$ and $\beta$ are zeros of a quadratic polynomial $4x^{2}+4x+1$, then find the quadratic polynomial whose zeros are $\alpha^{2} + \beta^{2}$ and $2\alpha\beta$.
To find the quadratic polynomial whose zeros are $\alpha^2 + \beta^2$ and $2\alpha\beta$, given that $\alpha$ and $\beta$ are the zeros of the polynomial $4x^2 + 4x + 1$, we will begin by using some properties and relationships between the coefficients and the zeros of the polynomial.
For a quadratic polynomial $ax^2 + bx + c$, the sum of the zeros $\alpha + \beta$ is given by: $$ \alpha + \beta = -\frac{b}{a} $$ Applying this to the given polynomial $4x^2 + 4x + 1$, we find: $$ \alpha + \beta = -\frac{4}{4} = -1 $$
The product of the zeros $\alpha \beta$ is given by: $$ \alpha \beta = \frac{c}{a} $$ Again, applying this to $4x^2 + 4x + 1$: $$ \alpha \beta = \frac{1}{4} $$
We need the expression for $\alpha^2 + \beta^2$, and this can be derived from the square of the sum of the roots: $$ (\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta $$ Substituting the known values: $$ (-1)^2 = \alpha^2 + \beta^2 + 2(\alpha \beta) = \alpha^2 + \beta^2 + 2\left(\frac{1}{4}\right) $$ $$ 1 = \alpha^2 + \beta^2 + \frac{1}{2} $$ Solving for $\alpha^2 + \beta^2$: $$ \alpha^2 + \beta^2 = \frac{1}{2} $$
Now, let's find a polynomial whose zeros are $\alpha^2 + \beta^2 = \frac{1}{2}$ and $2\alpha\beta = \frac{1}{2}$. We can use the polynomial format $x^2 - (\text{sum of roots})x + \text{product of roots}$: $$ x^2 - \left(\alpha^2 + \beta^2 + 2\alpha\beta\right)x + (\alpha^2 + \beta^2)(2\alpha\beta) $$ Plugging the values: $$ x^2 - \left(\frac{1}{2} + \frac{1}{2}\right)x + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) $$ $$ x^2 - x + \frac{1}{4} $$
Therefore, the quadratic polynomial whose zeros are $\alpha^2 + \beta^2$ and $2\alpha\beta$ is $\boldsymbol{x^2 - x + \frac{1}{4}}$.
Make a factor tree for the composite number 324.
To create a factor tree for the composite number 324, we start by finding factors of 324 that will help break it down into its prime factors. Here's how we can proceed:
First, recognize that 324 can be divided by 2 (since it's an even number). This gives us: $$ 324 = 2 \times 162 $$
Next, 162 is also even, and can similarly be divided by 2: $$ 162 = 2 \times 81 $$
Then, 81 is divisible by 3 (since the sum of digits, 8 + 1 = 9, is divisible by 3): $$ 81 = 3 \times 27 $$
Continuing, 27 can also be divided by 3: $$ 27 = 3 \times 9 $$
Finally, 9 is divisible by 3: $$ 9 = 3 \times 3 $$
At this point, we've reached all prime numbers. Organizing this into a factor tree:
Start with the composite number at the top:
324
Breaking down each composite factor until only prime factors are left:
324 splits into 2 and 162.
162 splits into 2 and 81.
81 splits into 3 and 27.
27 splits into 3 and 9.
9 splits into 3 and 3.
This gives us the prime factorization of 324 as: $$ 324 = 2^2 \times 3^4 $$
Here, 2 and 3 are the prime factors, and each number in the factor tree must eventually break down to these prime values.
If $r=3$ is a root of the quadratic equation $k r^{2}-k r-3=0$, then find the value of $k$.
Hello students! Today we are going to solve a problem where we are given a quadratic equation and we need to find the value of $k$. The equation provided is: $$ k r^2 - kr - 3 = 0 $$ We are also told that $r = 3$ is a root of this quadratic equation.
To find the value of $k$, we must substitute $r = 3$ into the equation, because it's given that when $r=3$, the equation should be equal to zero. Plugging in $r = 3$, we get: $$ k \cdot 3^2 - k \cdot 3 - 3 = 0 $$ Expanding this, we get: $$ 9k - 3k - 3 = 0 $$ Simplify by combining like terms: $$ 6k - 3 = 0 $$ Adding $3$ to both sides to solve for $k$, we have: $$ 6k = 3 $$ Dividing both sides by $6$, the calculation of $k$ is: $$ k = \frac{3}{6} = \frac{1}{2} $$
Therefore, the value of $k$ is $\frac{1}{2}$. Thank you for watching!
If $x=1$ is a root of the quadratic equation $2x^{2} - ax + 1$, then find the value of '$a$'.
To find the value of '$a$' in the quadratic equation $$2x^{2} - ax + 1 = 0$$ given that $x = 1$ is a root, we substitute $x = 1$ into the equation.
This substitution gives: $$ 2(1)^{2} - a(1) + 1 = 0 $$
Simplifying, we have: $$ 2 - a + 1 = 0 $$
Combining like terms, it results in: $$ 3 - a = 0 $$
Solving for '$a$', we find: $$ a = 3 $$
Thus, the value of '$a$' is 3.
If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 1 more than five times her actual age. What is her age now?
To solve the problem of finding Zeba's age, we start with the information that if Zeba were younger by 5 years, the square of her age then would have been 1 more than five times her current age.
Let's define Zeba's current age as $x$. This means that five years ago, her age would have been $x - 5$.
The problem statement translates to the following equation:
$$ (x - 5)^2 = 5x + 1 $$
Expanding the left side of the equation, we get:
$$ x^2 - 10x + 25 = 5x + 1 $$
To simplify this equation, bring all terms to one side:
$$ x^2 - 10x + 25 - 5x - 1 = 0 $$
Combine like terms:
$$ x^2 - 15x + 24 = 0 $$
This quadratic equation can be factored as:
$$ (x - 1)(x - 24) = 0 $$
So, the solutions for $x$ are:
$$ x = 1 \quad \text{or} \quad x = 24 $$
However, considering the physical context, if $x = 1$, then five years ago her age would have been $-4$, which is not possible. Therefore, the only practical and valid solution is:
$$ x = 24 $$
Thus, Zeba's current age is 24 years old.
Using the completing the square method, prove that the equation $x^{2} - 8x + 18 = 0$ has no real solution.
To solve the equation $x^2 - 8x + 18 = 0$ using the completing the square method, let's go through the steps to understand why it has no real solutions. Here's the breakdown:
Identify the coefficient of $x$, which is -8, and halve it, giving us $-4$.
Square this result: $(-4)^2 = 16$.
Rewrite the equation including this square and adjust the constant term to maintain equality. $$ x^2 - 8x + 16 = (x - 4)^2 $$ Notice, to keep the equation balanced, since you added 16 to the $x$ terms, you need to consider that with respect to the original constant (18), $$ (x-4)^2 + 2 $$ because $18 - 16 = 2$.
Now, the equation simplifies to: $$ (x-4)^2 + 2 = 0 $$
Rearrange to isolate the square term: $$ (x-4)^2 = -2 $$
Since the square of any real number is non-negative, $(x-4)^2$ must be $\geq 0$. Thus, there are no real values of $x$ for which $(x-4)^2 = -2$. This negative result indicates that it’s impossible for a real number square to equal a negative number.
Therefore, the equation $x^2 - 8x + 18 = 0$ has no real solutions. If real solutions were possible, $(x-4)^2$ would not result in a negative value.
Prove that is a right angle triangle, the square of the hypotenuse is equal the sum of the squares of other two sides.
Today, we will prove that in a right-angle triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides, as stated in the Pythagorean theorem.
Let's denote our right-angle triangle as ABC, where $ \angle B $ is $ 90^\circ $. We need to prove that: $$ AC^2 = AB^2 + BC^2 $$
To begin, let's drop a perpendicular from ( B ) to ( AC ), and call the point where it meets ( AC ), point ( D ).
From our construction and defining our angles, we can observe the following:
Both $ \triangle ADB $ and $ \triangle ABC $ have a common angle at $ A $ and both have a right angle. Thus, they share two corresponding angles: $ \angle A $ and the right angle.
By the AA (Angle-Angle) Similarity Postulate (two triangles with two corresponding angles being equal are similar), we conclude that $ \triangle ADB $ is similar to $ \triangle ABC $.
This similarity gives us the proportion between the sides of the triangles: $$ \frac{AB}{AC} = \frac{BC}{AB} $$ Cross-multiplying yields: $$ AB^2 = BC \cdot AC $$
Adding up the equations we derive from these similar triangles, and aligning them with our triangle properties, we find that: $$ AC^2 = AB^2 + BC^2 $$
Hence, we have proven that in a right-angled triangle, the square of the hypotenuse is the sum of the squares of the other two sides. This theorem is invaluable in geometry, and you'll find it useful in solving various problems involving right triangles.
What is the value of $\mathrm{k}$, if one of the zeroes of the quadratic polynomial $(k-2)x^{2} - 2x - 5$ is -1?
A) 5
B) 3
C) -5
D) 0
To determine the value of $k$ for which $-1$ is a zero of the quadratic polynomial $(k-2)x^2 - 2x - 5$, we can substitute $x = -1$ into the polynomial and set it equal to zero, because a zero of a polynomial is a value of $x$ that makes the polynomial equal to zero.
The given polynomial is: $$ (k-2)x^2 - 2x - 5. $$ Substitute $x = -1$: $$ (k-2)(-1)^2 - 2(-1) - 5 = 0. $$ Simplify the equation:
Since $(-1)^2 = 1$, the equation becomes: $$ (k-2)(1) + 2 - 5 = 0. $$
Further simplifying gives: $$ k - 2 + 2 - 5 = 0. $$
The $+2$ and $-2$ cancel out: $$ k - 5 = 0. $$ Finally, solving for $k$ gives $k = 5$.
Therefore, the value of $k$ is 5, which corresponds to option A.
If $\alpha$ and $\beta$ are the zeroes of the polynomial $f(x)=x^{2}-7x+12$, then find the value of $\frac{1}{\alpha}+\frac{1}{\beta}$.
A. 12
B. $-\frac{7}{12}$
C. -7
D. $\frac{7}{12}$
To find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$, where $\alpha$ and $\beta$ are the zeros of the polynomial $f(x) = x^2 - 7x + 12$, we start by using the relationship between the coefficients of the polynomial and its zeros:
For any quadratic polynomial $ax^2 + bx + c$, the sum and product of its zeros, $\alpha$ and $\beta$, can be found using:
Sum of the zeros ($\alpha + \beta$): $$ \alpha + \beta = -\frac{b}{a} $$
Product of the zeros ($\alpha \beta$): $$ \alpha \beta = \frac{c}{a} $$
For the given polynomial $x^2 - 7x + 12$, where $a = 1$, $b = -7$, and $c = 12$, we find:
$\alpha + \beta = -\frac{-7}{1} = 7$
$\alpha \beta = \frac{12}{1} = 12$
Now, using these, we aim to compute $\frac{1}{\alpha} + \frac{1}{\beta}$. This can be derived as: $$ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha \beta} = \frac{\alpha + \beta}{\alpha \beta} $$ Substituting the values we found: $$ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{7}{12} $$
Thus, the value of $\frac{1}{\alpha} + \frac{1}{\beta}$ is $\frac{7}{12}$, making the answer D. $\frac{7}{12}$.
For some integer $q$, every odd is of the form:
A) $m$
B) $m+1$
C) $2m$
D) $2m+1$
To determine the form that every odd integer can take, let's analyze each option given:
Option A: $q$
This option simply refers to an integer $q$ which can be either odd or even. Thus, representing all integers as simply $q$ does not guarantee they will be odd.Option B: $q+1$
Adding one to any integer $q$ will change its parity (from odd to even or even to odd). If $q$ is even, $q+1$ will be odd, and if $q$ is odd, $q+1$ will be even. Therefore, $q + 1$ does not uniformly represent all odd integers, as it could also be even depending on the value of $q$.Option C: $2q$
When we multiply an integer $q$ by 2, the result is definitely an even integer, regardless of whether $q$ is odd or even. Hence, $2q$ represents only even integers.Option D: $2q+1$
In this case, $2q$ is always even (as seen in Option C), so adding 1 to it makes $2q + 1$ always odd. Whether $q$ is odd or even, $2q$ is even, thus $2q + 1$ is an odd integer.
To conclusively illustrate with examples:
If $q = 1$, then $2q + 1 = 2 \times 1 + 1 = 3$ (odd)
If $q = 2$, then $2q + 1 = 2 \times 2 + 1 = 5$ (odd)
And, it continues similarly for all integer values of $q$.
The only option that guarantees an odd outcome for every integer $q$ is Option D: $2q+1$ which is then the correct choice for representing all odd integers.
Calculate the value of $k$, if $x=k$ is a solution of the quadratic polynomial $x^2 + 4x + 3$.
A. 1
B. -1
C. 3
D. -4
To find the value of $k$ such that $x=k$ is a solution for the quadratic polynomial $x^2 + 4x + 3$, we start by substituting $x$ with $k$ in the polynomial equation and setting it equal to zero: $$ k^2 + 4k + 3 = 0 $$
Next, we factorize this quadratic expression. We look for two numbers that multiply to $3$ (constant term) and add up to $4$ (coefficient of $k$). These numbers are $1$ and $3$. Therefore, we can rewrite the equation as: $$ (k + 1)(k + 3) = 0 $$
Setting each factor equal to zero gives the solutions for $k$: $$ k + 1 = 0 \implies k = -1 $$ $$ k + 3 = 0 \implies k = -3 $$
Among the provided options A, B, C, and D, where we have:
A. 1
B. -1
C. 3
D. -4
The correct answers according to our factorization are $k = -1$ and $k = -3$. However, only $k = -1$ is present in the options. Therefore, the answer is: Option B: -1.
Solve for $x$:
$$ x^2 - 11x + 28 = 0 $$
To solve the quadratic equation $x^2 - 11x + 28 = 0$, we can use the factoring method if the equation can be factored neatly. Here's how to proceed step-by-step:
Identify Coefficients: The quadratic equation is in the standard form $ax^2 + bx + c = 0$ where $a = 1$, $b = -11$, and $c = 28$.
Factoring: We look for two numbers whose product is equal to $a \times c$ (i.e., $1 \times 28 = 28$) and whose sum is equal to $b$ (i.e., $-11$).
Possible pairs for 28 are (1, 28) and (4, 7).
Among these, the numbers 4 and 7 sum up to 11, and we're looking for -11, so with negative signs the numbers are -4 and -7.
Write Factorization: We rewrite the quadratic equation using these numbers: $x^2 - 4x - 7x + 28 = 0$ Grouping the terms, we get: $(x^2 - 4x) - (7x - 28) = 0$
Factor Out Common Terms: From the first group, factor out $x$: $x(x - 4)$ From the second group, factor out -7: $-7(x - 4)$
Combine Like Terms: Now, the equation looks like: $x(x - 4) - 7(x - 4) = 0$ As both terms contain $(x - 4)$, factor it out: $(x - 7)(x - 4) = 0$
Set Each Factor Equal to Zero: The solutions occur where each factor equals zero: $$\begin{align} x - 7 &= 0 \quad \Rightarrow \quad x = 7 \\ x - 4 &= 0 \quad \Rightarrow \quad x = 4 \end{align}$$
Thus, the solutions are $x = 7$ and $x = 4$. These are the points where the equation $x^2 - 11x + 28 = 0$ is satisfied.
Find the sum of the roots of the equation $\left|x^{2}\right|-36|x|$
A) 36
B) -36
C) -1
D) 0
The correct option is D (0).
To solve for the sum of the roots, we start by recognizing that:
$$ \left| x^2 \right| = |x|^2 $$
The given equation is therefore:
$$ |x|^2 - 36|x| = 0 $$
We can factor out (|x|):
$$ |x|(|x| - 36) = 0 $$
This implies:
$$ |x| = 0 \quad \text{or} \quad |x| = 36 $$
So, solving for (x):
$$ x = 0 \quad \text{or} \quad x = \pm 36 $$
Thus, the roots are (0), (36), and (-36).
To find the sum of the roots, we add them:
$$ 0 + 36 - 36 = 0 $$
Therefore, the sum of the roots is 0.
Find the roots of $x^{2} - 32x - 900=0$.
To find the roots of the quadratic equation $ x^2 - 32x - 900 = 0 $, we use the method of middle term factorization. Here are the steps:
Identify the quadratic equation: $$ x^2 - 32x - 900 = 0 $$
Factorize the middle term: We need to split the middle term $-32x$ such that the factors multiply to give the constant term (-900).
We find two numbers whose product is (-900) and whose sum is $-32$. Those numbers are 50 and -18, because: $$ 50 \times -18 = -900 \quad \text{and} \quad 50 + (-18) = 32 $$
Rewrite the equation: Using these numbers, we rewrite the quadratic equation as: $$ x^2 - 50x + 18x - 900 = 0 $$
Group the terms and factorize: $$ (x^2 - 50x) + (18x - 900) = 0 $$ Factor out the greatest common factors in each group: $$ x(x - 50) + 18(x - 50) = 0 $$
Factor by grouping: Notice that $(x - 50)$ is a common factor: $$ (x - 50)(x + 18) = 0 $$
Solve for (x): Set each factor equal to zero: $$ x - 50 = 0 \quad \text{or} \quad x + 18 = 0 $$ Therefore, the solutions are: $$ x = 50 \quad \text{or} \quad x = -18 $$
The roots of the equation $ x^2 - 32x - 900 = 0$ are $\boxed{50}$ and $\boxed{-18}$.
If the equation $x^{2}+a x+b c=0$ and $x^{2}-b x+c a=0$ have a common root, then $a+b+c=$
A. 0
B. 1
C. -1
D. $3abc$
To solve this problem, we need to determine the value of $a + b + c$ given that the equations $x^2 + ax + bc = 0$ and $x^2 - bx + ca = 0$ share a common root.
Step-by-Step :
Given Equations:
$x^2 + ax + bc = 0$ (Equation 1)
$x^2 - bx + ca = 0$ (Equation 2)
Assume the common root is $\alpha$.
Substitute $\alpha$ into Equation 1: $$ \alpha^2 + a\alpha + bc = 0 \quad \text{(1)} $$
Substitute $\alpha$ into Equation 2: $$ \alpha^2 - b\alpha + ca = 0 \quad \text{(2)} $$
Subtract Equation 2 from Equation 1: $$ (\alpha^2 + a\alpha + bc) - (\alpha^2 - b\alpha + ca) = 0 $$ Simplifies to: $$ a\alpha + bc + b\alpha - ca = 0 $$
Rearrange terms to group similar factors: $$ (a + b)\alpha + bc - ca = 0 $$
This can be rewritten as: $$ (a + b)\alpha = ca - bc $$
Factor out ($ca - bc$) as: $$ (a + b)\alpha = c(a - b) \implies \alpha(a + b) = c(a - b) $$
Divide both sides by $(a - b)$: $$ \alpha = \frac{c(a - b)}{a + b} \quad \text{if} \ a + b \neq 0 $$
Substitute $\alpha = c$ into Equation 1: $$ c^2 + ac + bc = 0 $$
Combine terms: $$ c^2 + c(a + b) + bc = 0 $$
Factor out c: $$ c(c + a + b) + bc = 0 $$
Since this is a quadratic equation in $c$, the only solution that satisfies this equation uniformly is: $$ c(a + b + c) = 0 $$
As $c \neq 0$, we can conclude: $$ a + b + c = 0 $$
Conclusion:
We have successfully proved that $a + b + c = 0$.
Final Answer: $a + b + c = 0$
The greatest negative integer satisfying $x^2 - 4x - 77 < 0$ and $x^2 > 4$ is:
A. -2
B. -3
C. -1
D. -7
To find the greatest negative integer satisfying both inequalities, $x^2 - 4x - 77 < 0$ and $x^2 > 4$, you can follow these steps:
Step 1: Solve the Inequality $x^2 - 4x - 77 < 0$
We start by factoring the quadratic expression $x^2 - 4x - 77$.
Notice that $77 = 11 \times 7$ and $11 - 7 = 4$, which can help us factor the quadratic as follows: $$ x^2 - 4x - 77 < 0 $$ $$ (x - 11)(x + 7) < 0 $$
This results in two critical points: $x = 7$ and $x = -11$.
By using a number line to test values in the intervals defined by these critical points, we find that: $$ -11 < x < 7 $$
Step 2: Solve the Inequality $x^2 > 4$
This can be rewritten using the difference of squares: $$ x^2 - 4 > 0 $$ $$ (x - 2)(x + 2) > 0 $$
This results in two critical points: $x = 2$ and $x = -2$.
By analyzing these intervals, we find that: $$ x < -2 \quad \text{or} \quad x > 2 $$
Step 3: Combine s from Both Inequalities
The solution from the first inequality is $-11 < x < 7$.
The solution from the second inequality is $x < -2$ or $x > 2$.
Step 4: Find the Intersection
Visualizing the solutions on a number line, we need the intersection of these intervals: $$ \text{From } -11 < x < 7 \text{ and } x < -2 \quad \text{or} \quad x > 2: $$ $$ -11 < x < -2 \quad \text{or} \quad 2 < x < 7 $$
Step 5: Determine the Greatest Negative Integer in the Intersection
The interval $-11 < x < -2$ contains the range of negative integers less than -2.
The greatest negative integer within this interval is -3.
Conclusion
The greatest negative integer satisfying both given inequalities is -3.
Final Answer: B
The quadratic equation $a x^{2}+b x+c=0$ has two roots, then match the following lists.
Find the correct match from List-I to List-II.
List - I | List - II |
---|---|
A) both the roots are negative | 1) $ b = 0, ac < 0 $ |
B) both roots have opposite signs | 2) $ a > 0, b > 0, c > 0$ |
C) both roots are positive | 3) $ ac < 0$ |
D) both roots are equal in magnitude and opposite in sign | 4) $ b < 0, a > 0, c > 0 $ |
A $\begin{array}{llll} 3 & 2 & 4 & 1\end{array}$
B $\begin{array}{llll}2 & 3 & 4 & 1\end{array}$
C $\begin{array}{llll}3 & 2 & 1 & 4\end{array}$
D $\begin{array}{llll}2 & 1 & 3 & 4\end{array}$
To solve this question, we need to match the nature of the roots of the quadratic equation $ax^2 + bx + c = 0$ (List-I) with the conditions on the coefficients (a), (b), and (c) (List-II).
The quadratic equation has different roots based on the sign and values of its coefficients. Let's analyze each case from List-I and match it with List-II:
Analysis:
Both the roots are negative:
For both roots to be negative, the quadratic graph (parabola) must open upwards or downwards such that both roots lie on the negative side of the x-axis.
The vertex of the parabola $x = -\frac{b}{2a}$ should be negative.
This implies $\frac{b}{a}$ must be positive. Hence, $a$ and $b$ should have the same sign.
Therefore, $a > 0, b > 0, c > 0$ implies both roots are negative, and the correct match is $\textbf{2}$.
Both roots have opposite signs:
For roots to have opposite signs, the product of the roots (given by $\frac{c}{a}$) must be negative.
Therefore, $a$ and $c$ should have opposite signs.
This matches condition $ac < 0$, and the correct match is $\textbf{3}$.
Both roots are positive:
For both roots to be positive, again, the vertex condition $x = -\frac{b}{2a}$ should be positive.
This implies $\frac{b}{a}$ must be negative, which means $a$ and $b$ must have opposite signs.
This matches condition $b < 0, a > 0, c > 0$ implying both roots are positive, and the correct match is $\textbf{4}$.
Both roots are equal in magnitude and opposite in sign:
For roots to be equal in magnitude but opposite in sign, the sum of the roots (given by $-\frac{b}{a}$) should be zero. This implies $b = 0$.
Moreover, the product of the roots (given by $\frac{c}{a}$) must be negative, meaning that $a$ and $c$ should have opposite signs.
This matches condition $b = 0, ac < 0$, and the correct match is $\textbf{1}$.
Matching Lists:
A corresponds to (2) (both roots are negative).
B corresponds to (3) (both roots have opposite signs).
C corresponds to (4) (both roots are positive).
D corresponds to (1) (both roots are equal in magnitude and opposite in sign).
Final Answer:
The correct match is B: $ \begin{array}{llll}2 & 3 & 4 & 1\end{array} $
If $\alpha, \beta$ are the roots of the equation $x^2 - 2x + 3 = 0$, then the equation whose roots are $P = \alpha^3 - 3\alpha^2 + 5\alpha - 2$ and $Q = \beta^3 - \beta^2 + \beta + 5$ is
(A) $x^2 + 3x + 2 = 0$
(B) $x^2 - 5x + 4 = 0$
(C) $x^2 - 3x + 2 = 0$
(D) $x^2 + 5x + 4 = 0$
The correct option is C: $x^2 - 3x + 2 = 0$.
Given that $\alpha$ and $\beta$ are the roots of the equation: $$ x^2 - 2x + 3 = 0 $$
Therefore, we have: $$ \alpha^2 - 2\alpha + 3 = 0 \quad \cdots(1) $$ and $$ \beta^2 - 2\beta + 3 = 0 \quad \cdots(2) $$
From equation (1), rearranging terms gives: $$ \alpha^2 = 2\alpha - 3 $$ Cubing both sides: $$ \alpha^3 = \alpha \cdot \alpha^2 = \alpha (2\alpha - 3) = 2\alpha^2 - 3\alpha $$
We can now express $P$ as: $$ P = \alpha^3 - 3\alpha^2 + 5\alpha - 2 $$
Substituting the value of $\alpha^2$ from equation (1): $$ P = (2\alpha^2 - 3\alpha) - 3\alpha^2 + 5\alpha - 2 $$ $$ P = (2(2\alpha - 3) - 3\alpha) - 3(2\alpha - 3) + 5\alpha - 2 $$ Simplifying: $$ P = -\alpha^2 + 2\alpha - 2 $$ Using $\alpha^2 = 2\alpha - 3$: $$ P = 3 - 2 = 1 $$
Similarly for $\beta$: $$ Q = 2 $$
Sum and Product of the roots: $$ P + Q = 1 + 2 = 3 $$ $$ PQ = 1 \cdot 2 = 2 $$
Hence, the required quadratic equation with $P$ and $Q$ as roots is: $ x^2 - 3x + 2 = 0$
Thus, the correct option is C: $x^2 - 3x + 2 = 0$.
What must be subtracted from $3a^{2}-6ab-eb^{2}-1$ to get $4a^{2}-7ab-4b^{2}+1$?
To determine what must be subtracted from $3a^2 - 6ab - 3b^2 - 1$ to get $4a^2 - 7ab - 4b^2 + 1$, we need to find the difference between the two expressions.
First, represent the operation mathematically:
$$ (3a^2 - 6ab - 3b^2 - 1) - (x) = 4a^2 - 7ab - 4b^2 + 1 $$
To isolate $x$, rearrange the equation:
$$ x = (3a^2 - 6ab - 3b^2 - 1) - (4a^2 - 7ab - 4b^2 + 1) $$
Now, perform the subtraction:
Subtract each corresponding term from the first expression with the second expression:
$$ \begin{array}{rcl} (3a^2 - 4a^2) & = & -a^2 \ (-6ab - (-7ab)) & = & -6ab + 7ab = ab \ (-3b^2 - (-4b^2)) & = & -3b^2 + 4b^2 = b^2 \ (-1 - 1) & = & -2 \end{array} $$
Combine these results to form the final expression:
$$ x = -a^2 + ab + b^2 - 2 $$
So, the term that must be subtracted from $3a^2 - 6ab - 3b^2 - 1$ to get $4a^2 - 7ab - 4b^2 + 1$ is:
$$ \boxed{-a^2 + ab + b^2 - 2} $$
Sum of all the roots of the equation $x^{2} - 2x + |x-1| - 5 = 0$ is
A) 2
B) 1
C) 5
The correct option is A) 2.
Consider the given equation:
$x^{2} - 2x + |x-1| - 5 = 0$
We can rewrite this equation by substituting $ |x-1| $ as a separate variable:
$$ |x-1| = y $$
Thus, the equation becomes:
$$ y^2 + y - 6 = 0 $$
Next, we solve this quadratic equation:
$$ (y-2)(y+3) = 0 $$
Therefore, $y$ can be:
$$ |x-1| = 2 \quad \text{or} \quad |x-1| = -3 $$
Since the absolute value cannot be negative, we discard the $ |x-1| = -3 $ solution.
This leaves us with:
$$ |x-1| = 2 $$
So,
$$ x-1 = 2 \quad \text{or} \quad x-1 = -2 $$
This simplifies to:
$$ x = 3 \quad \text{or} \quad x = -1 $$
Summing all the roots:
$$ 3 + (-1) = 2 $$
Thus, the sum of all the roots of the equation is 2.
On dividing a quadratic polynomial by a linear polynomial, the quotient obtained is always:
a) a Constant
b) a linear polynomial
c) a quadratic polynomial
d) a cubic polynomial
When dividing a quadratic polynomial by a linear polynomial, the quotient obtained is always:
b) a linear polynomial
Justification:
A quadratic polynomial has the general form:
$$ ax^2 + bx + c $$
A linear polynomial has the form:
$$ dx + e $$
When you divide a quadratic polynomial by a linear polynomial, the degree of the quotient polynomial is reduced by 1. Since a quadratic polynomial has a degree of 2, the resulting quotient will have a degree of 1, which is a linear polynomial. Hence, the correct answer is b.
$2x^2 - 3y^2$ + $5x^2 + 6x^2$ + $-3x^2 - 4y^2$
Given expression:
$$ \begin{array}{l} (2x^2 - 3y^2) + (5x^2 + 6x^2) + (-3x^2 - 4y^2) \end{array} $$
First, simplify each term and then combine: $$ \begin{array}{l} 2x^2 - 3y^2 + 5x^2 + 6x^2 - 3x^2 - 4y^2 \end{array} $$
Next, group the like terms together (terms with $x^2$ and terms with $y^2$): $$ \begin{array}{l} (2x^2 + 5x^2 + 6x^2 - 3x^2) + (-3y^2 - 4y^2) \end{array} $$
Simplify the grouped terms: $$ \begin{array}{l} = (2 + 5 + 6 - 3)x^2 + (-3 - 4)y^2 \ = 10x^2 - 7y^2 \end{array} $$
Final simplified expression:$$ 10x^2 - 7y^2 $$
Note: Only like terms can be added or subtracted from each other.
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