Question

find the value of k for (k+4)x square+(k+1)x+1=0 has equal roots

Answer 1

Hey

Here is your answer,

(k+4)^2 +(k+1)x +1=0

b^2-4ac = (k+1)^2 - 4x (k+4)x 1
=k^2 +1 +2k -4k-16
= k^2 -2k -15

We have to solve the quadratic equation to find the value of k,

k^2 -2k -15=0
k^2+5k -3k -15=0
k(k+5)-3(k+5)=0
(k-3)(k+5)=0
k-3=9
K=3
K+5=0
K=-5

Hope it helps you!