1.

Find the values of k, for which the given equation has real roots:x2 + k(4x + k - 1) + 2 = 0

Answer»

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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