Q2 of 108 Page 44

Find that value of k in the following quadratic equation whose roots are real and equal:

kx(x – 2) + 6 = 0


Let us first solve the above equation,

kx2 – 2kx + 6 = 0


When we compare the above quadratic equation with the generalized one we get,


ax2 + bx + c = 0


a = k


b = -2k


c = 6


Since the quadratic equations have real and equal roots,


b2 – 4ac = 0 for real and equal roots


⟹ (-2k) 2 – (4 × k × 6) = 0


⟹ 4k2 – 24k = 0


⟹ 4k(k – 6) = 0


⟹ (k – 6) = 0 or 4k = 0


⟹ k = 6 or k = 0


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