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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: To find the value of $k$.
Solution:
Given –5 is a root of the quadratic equation $2x^{2} +p – 15 =0$
$\because $ –5 satisfies the given equation.
$\therefore 2( –5)^{2} + p( –5) – 15 = 0$
$\Rightarrow 50 – 5p – 15 =0$
$\Rightarrow 35 – 5p = 0$
$\Rightarrow \ 5p=35$
$\Rightarrow p=7$
Substituting $p=7$ in $p( x^{2} +x)+k=0$, we get
$7( x^{2} +x) +k=0$
$\therefore 7x^{2} +7x +k = 0$
The roots of the equation are equal.
$\therefore$ Discriminant $= b^{2} – 4ac = 0$
Here, $a =7$, $b=7$, $c = k$
$b^{2} -4ac=7^{2} -4\times 7\times k=49-28k=0$
$\Rightarrow 49=28k$
$\Rightarrow k=\frac{49}{28}$
$\Rightarrow k=\frac{7}{4}$
Hence, The value of $k=\frac{7}{4}$.
Updated on: 10-Oct-2022
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