Question

If the ratio of the roots of the equation $x^2-px+q=0,$ is $a:b,$ then the value of $p^{2}ab=$

• A. $qab$
• B. $q(a^2+b^2)$
• C. $q(a+b)$
• D. $q(a+b{)}^2$

Answer 1

The correct option is D $q(a+b{)}^2$

Let $\alpha ,\beta$ be the roots of equation $x^2-px+q=0$

$\therefore \alpha +\beta =p\cdots \left(i\right)\&\alpha \cdot \beta =q\cdots \left(ii\right)$

Since, the roots are of the ratio $a:b$
$\Rightarrow \frac{\alpha }{\beta }=\frac{a}{b}$

We have, $\frac{(\alpha +\beta {)}^2}{\alpha \beta }=\frac{{\alpha }^2+{\beta }^2+2\alpha \beta }{\alpha \beta }=\frac{{\alpha }^2}{\alpha \beta }+\frac{{\beta }^2}{\alpha \beta }+\frac{2\alpha \beta }{\alpha \beta }$

$\Rightarrow \frac{p^2}{q}=\frac{a}{b}+\frac{b}{a}+2$

$\Rightarrow \frac{p^2}{q}=\frac{a^2+b^2+2ab}{ab}$

$\Rightarrow \frac{p^2}{q}=\frac{(a+b{)}^2}{ab}$

$p^{2}ab=q(a+b{)}^2$