First slide

Theory of equations

Question

Let $α, β$ be the roots of $a x^{2} + b x + c = 0; γ, δ$ be the roots of $p x^{2} + q x + r = 0;$ and $D_{1}, D_{2}$ the respective discriminants of these equations. If $α, β, γ$ and $δ$ are in AP., then $D_{1} : D_{2} =$

Moderate

A

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

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

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

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

Solution

We have,
$α + β = - \frac{b}{a}, α β = \frac{c}{a}, γ + δ = - \frac{q}{p}$ and $γ δ = \frac{r}{p}$
$D_{1} = b^{2} - 4 a c$ and $D_{2} = q^{2} - 4 p r$
Now,
$α, β, γ, δ$ are in A.P.
$\begin{array}{l} \Rightarrow β - α = δ - γ \\ \Rightarrow (β - α)^{2} = (δ - γ)^{2} \\ \Rightarrow (β + α)^{2} - 4 α β = (γ + δ)^{2} - 4 γ δ \\ \Rightarrow \frac{b^{2}}{a^{2}} - \frac{4 c}{a} = \frac{q^{2}}{p^{2}} - \frac{4 r}{p} \end{array}$
$\Rightarrow \frac{b^{2} - 4 a c}{a^{2}} = \frac{q^{2} - 4 r p}{p^{2}} \Rightarrow \frac{D_{1}}{a^{2}} = \frac{D_{2}}{p^{2}} = \frac{D_{1}}{D_{2}} = \frac{a^{2}}{p^{2}}$

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