Question

If $-4$ is a root of the equation $x^2+px-4=0$ and the equation $x^2+px+q=0$ has equal root, find the values of $p$ and $q$.

Answer 1

$-4$ is the root of $x^2+px-4=0$

$\Rightarrow (-4{)}^2+p(-4)-4=0$

$\Rightarrow 16-4p-4=0$

$\Rightarrow 12=4p$

so $p=3$

$x^2+px-q=0$

$\Rightarrow x^2+3x+q=0$

It has equal roots so $D=0$

$b^2-4ae=0$

$9-4q=0$

so $q=\frac{9}{4}$

So $p=3$ & $q=\frac{9}{4}$.