Question

$-5$ is one of the zeroes of $2x^2+px-15$, zeroes of $p(x^2+x)+k$ are equal to each other. Find the value of $k$.

Answer 1

$-5$ is a root of quadratic equation $2x²+px-15=0$

so, $2(-5)²+p(-5)-15=0$

$=>50-5p-15=0$

$=>35-5p=0$

$p=7$

now, put $p=7$ in second quadratic equation,

$p(x²+x)+k=0$

$=>7(x²+x)+k=0$

$=>7x²+7x+k=0$

above equation has equal roots

so,$D=b²-4ac=0$

$=>7²-4\times 7\times k=0$

$=>7-4k=0$

$=>k=7/4=1.75$

hence, the value of $k=1.75$