Question

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

Answer 1

first we have to find for p.
$x = - 5 \\ 2( - 5) {}^{2} + p(5) - 15 = 0 \\ 50 + 5p - 15 = 0 \\ 5p = - 35 \\ p = \frac{ - 35}{5} \\ p = - 7$


p(x^2+x)+K = 0
-7 [ ( -5)^2 + (-5) ] + K =0
-7 ( 20) + K = 0
K = 140

Answer 2

We have,

2x²+px-15=0

and one root is -5

So,

Putting value of root in equation we get,

2(-5)² + p(-5) -15=0

2(25)- 5p- 15=0

50-15=5p

35/5 = P

P= 7

So, value of P is 7.

Again We have,

P(x²+x)+k= 0

Put value of P in this equation.

7(x²+x)+k=0

7x²+7x+k=0

We have given that, this equation has equal real roots.

So,

b² - 4ac= 0

7²-4(7)(K)= 0

49-28k = 0

28K = -49

k = -49/28

K = -7/4

So, value of k is -7/4.

$<marquee>Hope it helps You....</marquee>$