Q. If the equation $x^{2}+px+2q=0$ and $x^{2}+qx+2p=0\left(\right.p\neq q\left.\right)$ have a common root
then the absolute value of $\left(\right.p+q\left.\right)$ is

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A

B

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Solution:

We have,
$x^{2}+px+2q=0$
$x^{2}+qx+2p=0$
If $f\left(x\right)$ and $g\left(x\right)$ have common root then $f\left(x\right)\pmg\left(x\right)$ have same root common. Then,
$\left(x^{2} + p x + 2 q\right)-\left(x^{2} + q x + 2 p\right)=0$
$\Rightarrow x\left(\right.p-q\left.\right)-2\left(\right.p-q\left.\right)=0$
$\Rightarrow \left(x - 1\right)\left(p + q - 2\right)=0$
Hence, common root is
$p+q=-2$

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