Question

Find the values of k, for which the given equation has real roots:

x² + k(4x + k - 1) + 2 = 0

Answer 1

Roots are equal

∴ d = 0

d = b² – 4ac

d = (4k)² – 4 (k² – k + 2) (1)

d = 16k² – 4k² + 4k – 8

Put d = 0

0 = 16k² – 4k² + 4k – 8

12k² + 4k² + 4k – 8 = 0 (divide by 4)

3k² + k – 2 = 0

3k² + 3k – 2k – 2 = 0

3k (k + 1) – 2k (k + 1) = 0

(3k–2k)(k+1) = 0

(3k–2k)=0 or (k+1) = 0

k = 0 k = –1