Question

If the equation $x^2+px+q=0$ and $x^2+qx+p=0$ have exactly one root in common, then equation with other roots is

• A. $x^2-x-pq=0$
• B. $x^2-x+pq=0$
• C. $x^2+x-pq=0$
• D. $x^2+x+pq=0$

Answer 1

The correct option is D

$x^2+x+pq=0$

Given: $x^2+px+q=0$ and $x^2+qx+p=0$ have one root common.

Let the common root be $\alpha$.

$\Rightarrow {\alpha }^2+p\alpha +q=0$ and ${\alpha }^2+q\alpha +p=0$

Applying the condition of one root common,
$\frac{{\alpha }^2}{p^2-q^2}=\frac{\alpha }{q-p}=\frac{1}{q-p}$

$\Rightarrow \alpha =\frac{p^2-q^2}{q-p},1$

$\Rightarrow \frac{p^2-q^2}{q-p}=1$

$\Rightarrow p^2-q^2=q-p$

$\Rightarrow (p-q)[p+q+1]=0$

$\Rightarrow p-q=0$ and $p+q+1=0$

The required equation is $x^2-(p+q)x+pq=0$

$\Rightarrow$ $x^2+x+pq=0$