Q. 8 C5.0( 2 Votes )

# 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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