Question

15. Find the value of k for which the quadratic equation (k + 4)x2 + (k + 1)x + 1 = 0 has equal roots.

Answer 1

The given equation is:-
(k+4)x^2 + (k+1)x + 1=0

here ,
a= k+4
b= k+1
c=1

since , the equation have equal roots,
therefore, D= 0

==> b^2- 4ac = 0

==> (k+1)^2 - {4× (k+4)×1} = 0

==> k^2 + 1 + 2k -{ 4k + 16} = 0

==> k^2 + 1 + 2k - 4k - 16 = 0

==> k^2 - 2k - 15 = 0

==> k^2 - 5k + 3k - 15 = 0

==> k (k-5) + 3 (k-5) = 0

==> (k - 5)( k + 3)= 0

either of them =0

if k-5=0
then , k = 5

if k+3=0
then, k= -3

hence ,k = 5 or -3

hope this helps. ......

with regards...
#Misa