Q. 3 A4.0( 5 Votes )

# Find k, if the roots of x2 — (3k — 2)x + 2k = 0 are equal and real.

Comparing equation x2 — (3k — 2)x + 2k = 0 with ax2 + bx + c = 0 we get

a = 1, b = – (3k – 2) and c = 2k

Discriminant (D) = b2 – 4ac

As roots are real and equal D = 0

b2 – 4ac = 0

[ – (3k – 2)]2 – 4(1)(2k) = 0

(3k – 2)2 – 8k = 0

Expand using (a – b)2 = a2 – 2ab + b2

9k2 – 12k + 4 – 8k = 0

9k2 – 18k – 2k + 4 = 0

taking 9k common from first two terms and – 2 common from next two

9k(k – 2) – 2(k – 2) = 0

(9k – 2)(k – 2) = 0

9k – 2 = 0 or k – 2 = 0

9k = 2 or k = 2

Therefore k = and k = 2

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