Question

If \(\alpha,\beta\) are the roots of \(ax^2 + bx + c = 0\) and \(\alpha + k,~\beta + k\) are the roots of \(px^2 + qx + r = 0\).
Then \(\dfrac{b^2 - 4ac}{q^2 - 4pr}\) is equal to

Answer 1

Given: \(\alpha,\beta\) are the roots of \(ax^2 + bx + c = 0\) and \(\alpha + k,~\beta + k\) are the roots of \(px^2 + qx + r = 0\).

Absolute difference of both the roots will be same,
\(|\alpha-\beta|=|\alpha+k-\beta-k| \\ \Rightarrow [{(\alpha + k+\beta + k)}^2-4(\alpha + k)(\beta + k)]\\
= (\alpha + \beta)^2 - 4 \alpha \beta\\
\Rightarrow \dfrac{q^2}{p^2}-4 \dfrac{r}{p} = \dfrac{b^2}{a^2} - \dfrac{4c}{a}\\
\Rightarrow \dfrac{q^2 - 4pr}{p^2}=\dfrac{b^2 - 4ac}{a^2}\\
\Rightarrow \dfrac{b^2 - 4ac}{q^2 - 4pr}=\left ( \dfrac{a}{p} \right )^2\)