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# If $-5$ is a root of the quadratic equation $2x^2 + px -15 = 0$ and the quadratic equation $p(x^2 + x) + k = 0$ has equal roots, find the value of k.

**Given:**

$-5$ is a root of the quadratic equation $2x^2 + px -15 = 0$ and the quadratic equation $p(x^2 + x) + k = 0$ has equal roots.

**To do:**

We have to find the value of k.

**Solution:**

If $m$ is a root of the quadratic equation $ax^2+bx+c=0$ then it satisfies the given equation.

Therefore,

$2x^2 + px -15 = 0$

$2(-5)^2 + p(-5) -15 = 0$

$2(25)-5p-15=0$

$50-5p-15=0$

$35=5p$

$p=\frac{35}{5}$

$p=7$

Substituting the value of $p$ in $p(x^2 + x) + k = 0$, we get,

$7(x^2 + x) + k = 0$

$7x^2+7x+k=0$

Comparing the quadratic equation $7x^2+7x+k=0$ with the standard form of a quadratic equation $ax^2+bx+c=0$,

$a=7, b=7$ and $c=k$

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.

$D=(7)^2-4(7)(k)$

$D=49-28k$

The given quadratic equation has equal roots if $D=0$.

Therefore,

$49-28k=0$

$49=28k$

$k=\frac{49}{28}$

$k=\frac{7}{4}$

**The value of $k$ is $\frac{7}{4}$.**