Question

If the ratio of roots of the equation $x^2+px+q=0$ is equal to the ratio of roots of the equation $x^2+bx+c=0$, then $p^{2}c-b^{2}q=$

• A. $0$
• B. $2$
• C. $\frac{-3}{2}$
• D. $-2$

Answer 1

Answer: (A)

$0$

Let $\alpha ,\beta$ be the roots of equation $x^2+px+q=0$ and let $\gamma ,\delta$ be the roots of equation $x^2+bx+c=0$
Then, $\alpha +\beta =-p,\alpha \beta =q$
and $\gamma +\delta =-b,\gamma \delta =c$
Given, $\frac{\alpha }{\beta }=\frac{\gamma }{\delta }$
Applying componendo-dividendo
$\frac{\alpha +\beta }{\alpha -\beta }=\frac{\gamma +\delta }{\gamma -\delta }$
$\Rightarrow \frac{(\alpha +\beta {)}^2}{(\alpha -\beta {)}^2}=\frac{(\gamma +\delta {)}^2}{(\gamma -\delta {)}^2}$
$\Rightarrow \frac{p^2}{{\alpha }^2+{\beta }^2-2\alpha \beta }=\frac{b^2}{{\gamma }^2+{\delta }^2-2\gamma \delta }$
$\Rightarrow \frac{p^2}{(\alpha +\beta {)}^2-4\alpha \beta }=\frac{b^2}{(\gamma +\delta {)}^2-4\gamma \delta }$
$\Rightarrow \frac{p^2}{p^2-4q}=\frac{b^2}{b^2-4c}$
$\Rightarrow p^{2}c-b^{2}q=0$`
Hence, option 'A' is correct.