Question

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

Answer 1

Roots are equal, so discriminant is zero
${(k + 1)}^{2} - 4 \times (k + 4) \times 1 = 0 \\ {k}^{2} - 2k - 15 = 0 \\ (k - 5)(k + 3) = 0 \\ k = 5 \: \: or \: \: - 3$