Question

If the roots of the equation $x^2+px+q=0$ are equal to the roots of the equation $x^2+bx+c=0$ then prove that $p^{2}c=b^{2}q$

Answer 1

Let $\alpha ,\beta$ be roots of $x^2+px+q=0$
and $\gamma ,\delta$ be roots of $x^2+bx+c=0$
Now $\alpha \beta =q\alpha +\beta =-p$
let according to the question,
$\beta /\alpha =\frac{\delta }{\gamma }$..(1)
Now $\alpha +\beta =-p$
$\gamma +\delta =-b\frac{\alpha +\beta }{\gamma +\delta }=\frac{p}{b}$
$\frac{\alpha (1+\beta /\alpha )}{\gamma (1+\delta /\gamma )}=\frac{p}{b}$
$\frac{\alpha }{\gamma }=\frac{p}{b}(\beta /\alpha =\delta /\gamma )$ ..(2)
Now$\alpha \beta =q$
$\gamma \delta =c$
$\alpha \beta =q$
${\alpha }^2\beta /\alpha =q$
$\gamma \delta =c$
${\gamma }^2\delta /\gamma =c$\\
$\frac{{\alpha }^2\beta /\alpha }{{\gamma }^2\delta /\gamma }=\frac{q}{c}$
$\beta /\alpha =\delta /\gamma$
$\frac{\alpha }{\gamma }=\sqrt{\frac{q}{c}}$ ..(3)
from (2) and (3)
$\frac{p}{b}=\sqrt{\frac{q}{c}}$
$p^{2}c=b^{2}q$
Hence proved