Find the value of k such that the quadratic equation : x(x-2k)+6=0, has real and equal roots
The given quadratic equations is x(x – 2k) + 6 = 0.
x(x – 2k) + 6 = 0
∴ x2 – 2kx + 6 = 0
Discriminant of the given equation, D = b2 – 4ac = (– 2k)2 – 4 × 1 × 6 = 4k2 – 24
Given, roots of the given quadratic equation are real and equal.
∴ D = 0
⇒ 4k2 – 24 = 0
⇒ k2 = 6
Thus, the value of k is 
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