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

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(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|

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

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 $\left| \frac{a}{A} \right| = \sqrt{\frac{D_1}{D_2}}$.

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