Question:
If $-5$ is a root of the quadratic equation $2 x^{2}+p x-15=0$ and the quadratic equation $p\left(x^{2}+x\right)+k=0$ has equal roots, find the value of $k$.
Solution:
It is given that $-5$ is a root of the quadratic equation $2 x^{2}+p x-15=0$.
$\therefore 2(-5)^{2}+p \times(-5)-15=0$
$\Rightarrow-5 p+35=0$
$\Rightarrow p=7$
The roots of the equation $p x^{2}+p x+k=0=0$ are equal.
$\therefore D=0$
$\Rightarrow p^{2}-4 p k=0$
$\Rightarrow(7)^{2}-4 \times 7 \times k=0$
$\Rightarrow 49-28 k=0$
$\Rightarrow k=\frac{49}{28}=\frac{7}{4}$
Thus, the value of $k$ is $\frac{7}{4}$.