Q8 of 168 Page 7

(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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