If $- 5$ is one root of quadratic equation $2 x^{2} + p x - 15 = 0$ and roots of quadratic equation $p (x^{2} + x) + k = 0$ is equal ,then find the value of $k$ .

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Solution

Given,
$p (x^{2} + x) + k = 0$
has equal roots
so, $b^{2} = 4 a c$
$p x^{2} + p x + k = 0$
$(p)^{2} = 4 (p) (k)$
$p = 0, o r p = 4 k$
and given -5 is root of $2 x^{2} + p x - 15 = 0$
put $x = - 5$
$2 (- 5)^{2} + (- 5) p - 15 = 0$
$50 - 5 p - 15 = 0$
$p = 7$
$7 = 4 k$
$k = 7 / 4$

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Similar questions

Q. If -5 is a root of the quadratic equation

2 x^{2} + p x - 15 = 0

and the quadratic equation

p (x^{2} + x) + k = 0

has equal roots, find the value of k.

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

Q. If

- 4

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2 x^{2} + p x + 8 = 0

and

p (x^{2} + x) + k = 0

then find the value of

k

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If −5 is one root of quadratic equation 2x2+px−15=0 and roots of quadratic equation p(x2+x)+k=0 is equal ,then find the value of k.

Given, p(x2+x)+k=0 has equal roots so,b2=4ac px2+px+k=0 (p)2=4(p)(k) p=0,orp=4kand given -5 is root of 2x2+px−15=0put x=−52(−5)2+(−5)p−15=050−5p−15=0p=77=4kk=7/4

If $- 5$ is one root of quadratic equation $2 x^{2} + p x - 15 = 0$ and roots of quadratic equation $p (x^{2} + x) + k = 0$ is equal ,then find the value of $k$ .

Given,
$p (x^{2} + x) + k = 0$
has equal roots
so, $b^{2} = 4 a c$
$p x^{2} + p x + k = 0$
$(p)^{2} = 4 (p) (k)$
$p = 0, o r p = 4 k$
and given -5 is root of $2 x^{2} + p x - 15 = 0$
put $x = - 5$
$2 (- 5)^{2} + (- 5) p - 15 = 0$
$50 - 5 p - 15 = 0$
$p = 7$
$7 = 4 k$
$k = 7 / 4$