Q11 of 168 Page 7

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

x2 + k(4x + k — 1) + 2 = 0

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


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