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

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

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

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

None of these

answer is B.

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Detailed Solution

It is given that the ratio of the roots of the equation

a x^{2} + bx + c = 0 is

same as the ratio of the roots of the equation

p x^{2} + qx + r = 0

.
Let

α_{1}, β_{1}

be the roots of

a x^{2} + bx + c = 0

and

α_{2}, β_{2}

be the roots of

p x^{2} + qx + r = 0

.
We can write according to the question that,

\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 and dividendo)

\Rightarrow \frac{{(α_{1} + β_{1})}^{2}}{{(α_{1} - β_{1})}^{2}} = \frac{{(α_{2} + β_{2})}^{2}}{{(α_{2} - β_{2})}^{2}}

⇒

\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{\frac{b^{2}}{a^{2}}}{\frac{b^{2} - 4 ac}{a^{2}}} = \frac{\frac{q^{2}}{p^{2}}}{\frac{q^{2} - 4 rp}{p^{2}}}

\Rightarrow \frac{b^{2}}{D_{1}} = \frac{q^{2}}{D_{2}}

\Rightarrow \frac{D_{1}}{D_{2}} = \frac{b^{2}}{q^{2}}

Hence, the correct option is 2.

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