$\begin{matrix} If α, β are the roots of a x^{2} + b x + c and α + h, β + h are the roots of p x^{2} + q x + r = 0 and D_{1}, D_{2} are the respective discriminants of these equations, then D_{1} : D_{2} = \end{matrix}$

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

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

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

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Solution

The correct option is A $\frac{a^{2}}{p^{2}}$
Let $A = α + h$ and $B = β + h$

Then, $A - B = (α + h) - (β + h)$

$\Rightarrow A - B = α - β$

$\Rightarrow {(A - B)}^{2} = {(α - β)}^{2}$

$\Rightarrow {(A + B)}^{2} - 4 A B = {(α + β)}^{2} - 4 α β$

$\Rightarrow \frac{b^{2}}{a^{2}} - \frac{4 c}{a} = \frac{q^{2}}{p^{2}} - \frac{4 r}{p}$

$\Rightarrow \frac{b^{2} - 4 a c}{a^{2}} = \frac{q^{2} - 4 r p}{p^{2}}$

$\Rightarrow \frac{D_{1}}{a^{2}} = \frac{D_{2}}{p^{2}}$

$\Rightarrow \frac{D_{1}}{D_{2}} = \frac{a^{2}}{p^{2}}$

Hence $D_{1} : D_{2} = a^{2} : p^{2} = \frac{a^{2}}{p^{2}}$

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$α, β$ are the roots of $a x^{2} + b x + c = 0 a n d γ, δ a r e t h e r o o t s o f p x^{2} + q x + r = 0 a n d D_{1}, D_{2}$ be the respective discriminants of these equations. If $α, β, γ, a n d δ$ are in A.P. then $D_{1} : D_{2} = (w h e r e α, β, γ δ, ϵ R & a, b, c, p, q, r ϵ R)$