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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 $\frac{7}{4}$.
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