Find the value of k for which the equation kx[x-2]+6=0 has equal roots.

@Himanshu4: Good answer! Keep posting!

The given equation is *kx *(*x* – 2) + 6 = 0.

*kx* (*x* – 2) + 6 = 0

∴ *kx*^{2} – 2*kx* + 6 = 0

Here *a *= *k*, *b* = – 2*k *and *c* = 6.

Discriminant of the quadratic equation = *D* = *b*^{2} – 4*ac* = (–2*k*)^{2} – 4 × *k* × 6 = 4*k*^{2} – 24*k*

Roots of the given quadratic equation are equal.

∴* D *= 0

⇒ 4*k*^{2} – 24*k* = 0

⇒ 4*k*(*k* – 6) = 0

⇒ *k* = 0 or *k* – 6 = 0

⇒ *k* = 0 or *k* = 6

∴*k* = 6 (∵ *k* ≠ 0)

Thus, the value of *k* is 6.

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