(i) The given quadratic equation is
Kx^{2} + 6x – 3k = 0
Here, a = k, b = 6 and c = - 3k
Sum of the roots α + β = -b/a = - 6/k
And product of the roots αβ = c/a = -3k/k
Since, Sum of the roots = product of the roots
⇒ - 6/k = -3
⇒ k = + 6/+ 3 ⇒ k = 2
(ii) The given equation is
(k + 1)x^{2} + (2k + 1)x – 9 = 0
Here, a = k + 1, b = (2k + 1) and c = - 9
Sum of the roots α + β = -(2k + 1)/k + 1
and αβ = c/a = - 9/k + 1
Since, sum of the roots = Product of the roots
Then, (2k + 1/k + 1) = 9/k + 1
⇒ 2k + 1 = 9
⇒ 2k = 9 – 1
⇒ 2k = 8
⇒ k = 8/2 = 4
⇒ k = 4