Question

If the difference between the roots of $x^2+2px+q=0$ be same as that between the roots of $x^2+2qx+p=0$ $(p\ne q)$, then $p+q$ is equal to

Answer 1

Let $\alpha$ and $\beta$ are the roots of the equation $x^2+2px+q=0$ and ${\alpha }^{\prime }$ and ${\beta }^{\prime }$ be the roots of the equation $x^2+2qx+p=0$.

$\alpha -\beta =\alpha {}^{\prime }-\beta {}^{\prime }\dots \left[Given\right]$

$\Rightarrow \sqrt{{(\alpha +\beta )}^2-4\alpha \beta }=\sqrt{{({\alpha }^{\prime }+{\beta }^{\prime })}^2-4\alpha {}^{\prime }\beta {}^{\prime }}$

$\Rightarrow \sqrt{{(-2p)}^2-4q}=\sqrt{{(-2q)}^2-4p}$

$\Rightarrow 4p^2-4q^2=4q-4p$

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

$\Rightarrow (p-q)(p+q)=-(p-q)$

$\Rightarrow p+q=-1$

Hence, value of $p+q$ is equal to $-1$.