First slide

Theory of equations

Question

The ratio of the roots of the equation $a x^{2} + b x + c = 0$ is same as the ratio of the roots of the equation $p x^{2} + q x + r = 0 .$ If $D_{1}$ and $D_{2}$ are the discriminants of $a x^{2} + b x + c = 0$ and $p x^{2} + q x + r = 0$ respectively, then $D_{1} : D_{2} =$

Moderate

A

$\frac{a^{2}}{p^{2}}$
B

$\frac{b^{2}}{q^{2}}$
C

$\frac{c^{2}}{r^{2}}$
D

none of these

Solution

Let $α_{1}, β_{1}$ be the roots of $a x^{2} + b x + c = 0$ and $α_{2}, β_{2}$ be the roots of $p x^{2} + q x + r = 0 .$ Then,
$\frac{α_{1}}{β_{1}} = \frac{α_{2}}{β_{2}}$ [Given]
$\Rightarrow \frac{α_{1} + β_{1}}{α_{1} - β_{1}} = \frac{α_{2} + β_{2}}{α_{2} - β_{2}}$ [Applying componendo land dividendo]
$\begin{array}{l} \Rightarrow \frac{{(α_{1} + β_{1})}^{2}}{{(α_{1} - β_{1})}^{2}} = \frac{{(α_{2} + β_{2})}^{2}}{{(α_{2} - β_{2})}^{2}} \\ \Rightarrow \frac{{(α_{1} + β_{1})}^{2}}{{(α_{1} + β_{1})}^{2} - 4 α_{1} β_{1}} = \frac{{(α_{2} + β_{2})}^{2}}{{(α_{2} + β_{2})}^{2} - 4 α_{2} β_{2}} \\ \Rightarrow \frac{b^{2} / a^{2}}{\frac{b^{2} - 4 a c}{a^{2}}} = \frac{q^{2} / p^{2}}{\frac{q^{2} - 4 r p}{p^{2}}} \Rightarrow \frac{b^{2}}{D_{1}} = \frac{q^{2}}{D_{2}} \Rightarrow \frac{D_{1}}{D_{2}} = \frac{b^{2}}{q^{2}} \end{array}$

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