If roots of the equation ax^2 + bx + c = 0 ar | vedclass

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

Question

(C) For the equation $ax^2 + bx + c = 0$,the roots are $r_1 = \alpha - \beta$ and $r_2 = \gamma - \delta$. The difference of the roots is $|r_1 - r_2|

Vedclass · Accepted Answer

$\sqrt{\frac{D_1}{D_2}}$

Vedclass · Answer

(C) For the equation $ax^2 + bx + c = 0$,the roots are $r_1 = \alpha - \beta$ and $r_2 = \gamma - \delta$. The difference of the roots is $|r_1 - r_2| = |(\alpha - \beta) - (\gamma - \delta)| = |\alpha - \beta - \gamma + \delta| = \frac{\sqrt{D_1}}{|a|}$.For the equation $Ax^2 + Bx + C = 0$,the roots are $R_1 = \alpha + \delta$ and $R_2 = \beta + \gamma$. The difference of the roots is $|R_1 - R_2| = |(\alpha + \delta) - (\beta + \gamma)| = |\alpha - \beta + \delta - \gamma| = \frac{\sqrt{D_2}}{|A|}$.Since $|\alpha - \beta - \gamma + \delta| = |\alpha - \beta + \delta - \gamma|$,we have $\frac{\sqrt{D_1}}{|a|} = \frac{\sqrt{D_2}}{|A|}$.Therefore,$\left| \frac{a}{A} \right| = \sqrt{\frac{D_1}{D_2}}$.

Similar Questions

If $\alpha$ and $\alpha^2$ are the roots of the equation $x^2 + x + 1 = 0$,then the equation whose roots are $\alpha^{31}$ and $\alpha^{62}$ is .....

If $2, 1, 1$ are roots of the equation $x^3-4x^2+5x-2=0$,then find the roots of the equation $\left(x+\frac{1}{3}\right)^3-4\left(x+\frac{1}{3}\right)^2+5\left(x+\frac{1}{3}\right)-2=0$.

The number of real roots of the equation $x|x-2|+3|x-3|+1=0$ is:

If $a + b + c = 0$,$a \ne 0$,and $a, b, c \in \mathbb{Q}$,then both the roots of the equation $ax^2 + bx + c = 0$ are

If $x = \sqrt{7} + \sqrt{3}$ and $xy = 4$,then find the value of $x^4 + y^4$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module