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

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Updated on: 10-Oct-2022

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