Question

If $x^2-ax+b=0$ and $x^2-px+q=0$ have one root common and the second equation has equal roots, then $b+q$ is equal to

• A. $p+a$
• B. $\frac{p}{2}$
• C. $\frac{ap}{2}$
• D. $\frac{a}{2}$

Answer 1

The correct option is C $\frac{ap}{2}$

Let the root of $x^2-px+q=0$ be $\beta$
$\Rightarrow 2\beta =p,{\beta }^2=q$
$\Rightarrow \beta =\frac{p}{2}=\pm \sqrt{q}\cdots \left(1\right)$

Now for $x^2-ax+b=0$,
let the roots be $\alpha$ and $\beta$.
$\alpha +\beta =a,\alpha \beta =b$
From $\left(1\right)$,
$\alpha =\frac{2a-p}{2},\frac{2a-p}{2}\cdot \frac{p}{2}=b$

$\frac{p}{2}\frac{(2a-p)}{2}=b$
$\Rightarrow 4b=2ap-p^2$
$\Rightarrow 4b+p^2=2ap$
$\Rightarrow 4b+4q=2ap[\because \frac{p}{2}=\pm \sqrt{q}]$
$\Rightarrow b+p=\frac{ap}{2}$