Question

–5 is one of the zeroes of 2x² +px - 15, zeroes of p(x² + x) + k are equal to each other. Find the value of k.

Answer 1

Given:
-5 is a root of 2x²+px-15=0

Put x= -5
2(-5)²+p(-5)-15=0
2×25 -5p -15=0
50 -5p -15= 0
35 -5p = 0
35 = 5p
p = 35/5
p = 7

Now p(x²+x)+ k = 0
7(x²+x)+ k = 0    [p=7]
7x²+7x+ k = 0………(1)

On comparing equation 1 with ax²+bx+c =0
a= 7, b= 7, c= k

Discriminant (D) = b² -4ac
=7² - 4× 7 × k
= 49 -28k

Since, p(x²+x)+ k = 0 has equal roots.
i.e 7x²+7x+ k = 0 has equal roots.

D= 0
49 -28k = 0
28k = 49
k = 49/28
k= 7/4

Hence, required value of k = 7/4.

HOPE THIS WILL HELP YOU....