Find the value of k for which the quadratic equation (k + 1)x² – 6(k + 1)x + 3(k + 9) = 0, k ≠ – 1 has equal roots.

Given: Equation (k + 1)x² – 6(k + 1)x + 3(k + 9) = 0 has equal roots

To find: Value of k

Concept used:

If b² – 4ac = 0 then roots of the quadratic equation ax² + bx + c = 0 are equal

Since the roots are equal,

b² – 4ac = 0 …(i)

Here, a = (k + 1), b = -6(k +1) and c = 3(k + 9)

Putting the values in eq. (i), we get

[-6(k +1)]² – 4(k +1)[3(k + 9)] = 0

⇒ 36(k +1)² – 12 (k + 1)(k + 9) = 0

Taking 12(k + 1) common, we get

⇒ 12(k + 1)[3(k +1) – (k + 9)] = 0

⇒ (k + 1) [3k + 3 – k – 9] = 0

⇒ (k + 1) (2k – 6) = 0

⇒ k + 1 = 0

⇒ k = -1

but k = -1 is not possible as it is given k ≠ -1

⇒ 2k – 6 = 0

⇒ 2k = 6

⇒ k = 3

Hence, the value of k is 3