If alpha and beta are the roots of the equati | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given: $\alpha$ and $\beta$ are roots of the equation $ax^2+bx+c=0$. Sum of roots: $\alpha+\beta = -\frac{b}{a}$. Product of roots: $\alpha\beta =

Vedclass · Accepted Answer

$acx^2+(ab+bc)x+b^2=0$

Vedclass · Answer

(C) Given: $\alpha$ and $\beta$ are roots of the equation $ax^2+bx+c=0$. Sum of roots: $\alpha+\beta = -\frac{b}{a}$. Product of roots: $\alpha\beta = \frac{c}{a}$. New roots are $S_1 = \alpha+\beta = -\frac{b}{a}$ and $S_2 = \frac{1}{\alpha}+\frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{-b/a}{c/a} = -\frac{b}{c}$. The required equation is $(x-S_1)(x-S_2) = 0$. $(x - (-\frac{b}{a}))(x - (-\frac{b}{c})) = 0$. $(x + \frac{b}{a})(x + \frac{b}{c}) = 0$. Multiplying by $ac$: $(ax+b)(cx+b) = 0$. $acx^2 + abx + bcx + b^2 = 0$. $acx^2 + (ab+bc)x + b^2 = 0$.

If $\alpha$ and $\beta$ are the roots of the equation $ax^2+bx+c=0$,then the equation whose roots are $\alpha+\beta$ and $\frac{1}{\alpha}+\frac{1}{\beta}$ is

Similar Questions

Let $r_1, r_2, r_3$ be roots of the equation $x^3 - 2x^2 + 4x + 5074 = 0$. Then the value of $(r_1 + 2)(r_2 + 2)(r_3 + 2)$ is

The equation whose roots are the reciprocals of the roots of the equation $3x^2 - 20x + 17 = 0$ is

If $\alpha$ and $\beta$ are the roots of the equation $4x^2 - \sqrt{13}x - 7 = 0$,then what is the value of $|\alpha - \beta|$?

If the roots of the equation $(x - a)(x - b) = c$ (where $c \neq 0$) are $\alpha$ and $\beta$,then what are the roots of the equation $(x - \alpha)(x - \beta) + c = 0$?

The roots of the equation $x^2 + ax + b = 0$ are $p$ and $q$. Then the equation whose roots are $p^2q$ and $pq^2$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module