If the roots of the equation ax^2 + bx + c = | 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 $x_1 = \alpha - \beta$ and $x_2 = \gamma - \delta$.The difference of the roots is $|x_1 - x_2|

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) For the equation $ax^2 + bx + c = 0$,the roots are $x_1 = \alpha - \beta$ and $x_2 = \gamma - \delta$.The difference of the roots is $|x_1 - x_2| = |(\alpha - \beta) - (\gamma - \delta)| = |\alpha - \beta - \gamma + \delta| = \frac{\sqrt{D_1}}{|a|}$,where $D_1 = b^2 - 4ac$.For the equation $Ax^2 + Bx + C = 0$,the roots are $y_1 = \alpha + \delta$ and $y_2 = \beta + \gamma$.The difference of the roots is $|y_1 - y_2| = |(\alpha + \delta) - (\beta + \gamma)| = |\alpha - \beta + \delta - \gamma| = \frac{\sqrt{D_2}}{|A|}$,where $D_2 = B^2 - 4AC$.Since $|\alpha - \beta - \gamma + \delta| = |\alpha - \beta + \delta - \gamma|$,the absolute differences of the roots are equal.Therefore,$\frac{\sqrt{D_1}}{|a|} = \frac{\sqrt{D_2}}{|A|}$.Rearranging the terms,we get $\left| \frac{a}{A} \right| = \sqrt{\frac{D_1}{D_2}}$.

Similar Questions

If $l, m, n$ are real and $l \ne m$,then the roots of the equation $(l - m)x^2 - 5(l + m)x - 2(l - m) = 0$ are

If the ratio of the roots of $ax^2 + 2bx + c = 0$ is the same as the ratio of the roots of $px^2 + 2qx + r = 0$,then

If $\frac{2x}{2x^2 + 5x + 2} > \frac{1}{x + 1}$,then

The equation formed by decreasing each root of $ax^2 + bx + c = 0$ by $1$ is $2x^2 + 8x + 2 = 0$. Then:

If one root of $ax^2 + bx + c = 0$ is the square of the other,then the value of $b^3 + ac^2 + a^2c$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module