Find the value of k for which the equation x2 + k(2x + k − 1)

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354.

Find the value of k for which the equation x2 + k(2x + k − 1) + 2 = 0 has real and equal roots. 


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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