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Question
Find the value of k for which the equation x2 + k(2x + k − 1) + 2 = 0 has real and equal roots.
Solution
The given equation is x2+k(2x+k−1)+2=0.
⇒x2+2kx+k(k−1)+2=0
So, a = 1, b = 2k, c = k(k − 1) + 2
We know D=b2−4ac
⇒D=(2k)2 − 4 × 1 × [k(k − 1) + 2]
⇒D=4k2 − 4[k2 − k + 2]
⇒D=4k2 − 4k2 + 4k − 8
⇒D=4k − 8 = 4(k − 2)
For equal roots, D = 0
Thus, 4(k − 2) = 0
So, k = 2.
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