Q8 of 47 Page 1

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


Simplifying the equation: x2 + 2kx + (k2 – k + 2) = 0


Roots of an equation are given by the formula:


x = , where, D = b2 – 4ac


For equal roots, D = 0


On comparing with ax2 + bx + c = 0, we get:


a = 1, b = 2k , c = (k2 – k + 2)


Therefore, D = 0


4k2 – 4(k2 – k + 2) = 0


⇒ 4k2 – 4k2 + 4k - 8 = 0


⇒ 4k – 8 = 0


⇒ 4k = 8


⇒ k = 2


∴ k = 2


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