Q. 8 B5.0( 3 Votes )

# (k + 4)x2 + (k + 1)x + 1 = 0

Since roots are equal

d=0 (1)

(k + 4)x2 + (k + 1)x + 1 = 0

d=b2–4ac

d = (k – 1)2– 4 (k + 4) (1)

d = (–2k + 2)2 – 4k – 4

d = k2 + 1+ 2k – 4k – 16

From (1), d = 0

Equation will be:

0 = k2 + 1 + 2k – 4k – 16

k2 – 2k – 15 = 0

k2 – 5k + 3k – 15 = 0

k(k – 5) + 3 (k – 5) = 0

(k – 5) (k + 3) = 0

K – 5 = 0 k + 3 = 0

k = 5 k = –3

Values of k are –3, 5.

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