Q8 of 168 Page 7

k2x2 — 2(2k — 1)x + 4 = 0

Since roots are equal


∴ d=0 (1)


k2x2 — 2(2k — 1)x + 4 = 0


d = b2 – 4ac


d = (–2(k–1))2 – 4 (k2) (4)


d = (–2k+2)2 – 4k – 4


d = (–4k+2)2 – 16k2


(∵ (a – b)2 = a2 + b2 – 2ab)


d = 16k2 – 16k + 4 – 16k2


d = –16k + 4


From (1), d = 0


∴ Equation will be:


0 = –16k + 4


16k = 4




∴ Values of k are is .


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