Question

Find the value of k for which the equation x^{​​​​​​2}+k(2x+k-1)+2 has real and equal roots.
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Answer 1

Given, equation, x^2 + K (2x + K – 1) + 2 = 0

x^2 + 2Kx + K(K – 1) + 2 = 0

For real and equal roots, D = 0

⇒ D = b^2 – 4ac = 0

⇒ (2K)^2 – 4 [1] [ K(K – 1) + 2 ] = 0

4K^2 – 4 [ K^2 – K + 2 ] = 0

4K^2 – 4K^2 + 4K – 8 = 0

4K – 8 = 0

4K =8

K=8/4

=2

K=2