Question

Find the value of k ,if (k+4)x2+ (k+1)x +3 = 0 has equal roots.

Answer 1

$answer$

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

D=b2 -4ac =(k+1)2-4(k+4)(1)

=k2 +2k+1-4k-16

=k2 -2k-15

For equal roots, D=0

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=0 OR k-5=0

k = -3, k = 5.

Answer 2

Answer:

$your \: ans\: is \: k = 5 \: and \: k = - 3$