Find the values of k for which each of the following quadratic equation has equal roots :

(i) x² + 4kx + (k² - k + 2) = 0

(ii) (k - 4)x² + 2(k - 4)x + 4 = 0

(i) The given equation is x² + 4kx + (k² - k + 2) = 0.

Comparing with ax² + bx + c = we obtain,
a = 1 , b = 4k , c = (k² - k + 2)

$\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (4k)^2 - 4 \times 1 \times (k^2 - k + 2) \\[1em] = 16k^2 - 4(k^2 - k + 2) \\[1em] = 16k^2 - 4k^2 + 4k - 8 \\[1em] = 12k^2 + 4k - 8$

For equal roots, discriminant = 0

$\Rightarrow 12k^2 + 4k - 8 = 0 \\[1em] \Rightarrow 12k^2 + 12k - 8k - 8 = 0 \\[1em] \Rightarrow 12k(k + 1) - 8(k + 1) = 0 \\[1em] \Rightarrow (k + 1) - (12k - 8) = 0 \\[1em] \Rightarrow (k + 1)(12k - 8) = 0 \\[1em] \Rightarrow k + 1 = 0 \text{ or } 12k - 8 = 0 \\[1em] \Rightarrow k = -1 \text{ or } k = \dfrac{8}{12} \\[1em] \Rightarrow k = -1 \text{ or } k = \dfrac{2}{3}$

Hence, the value of k is -1, $\dfrac{2}{3}$.

(ii) The given equation is (k - 4)x² + 2(k - 4)x + 4 = 0.

Comparing with ax² + bx + c = we obtain,
a = k - 4 , b = 2(k - 4) , c = 4

$\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (2k - 8)^2 - 4 \times k - 4 \times 4 \\[1em] = 4k^2 + 64 - 32k - 16(k - 4) \\[1em] = 4k^2 - 32k - 16k + 64 + 64 \\[1em] = 4k^2 - 48k + 128$

For equal roots, discriminant = 0

$\Rightarrow 4k^2 - 48k + 128 = 0 \\[1em] \Rightarrow 4(k^2 - 12k + 32) = 0 \\[1em] \Rightarrow k^2 - 12k + 32 = 0 \\[1em] \Rightarrow k^2 - 8k - 4k + 32 = 0 \\[1em] \Rightarrow k(k - 8) - 4(k - 8) = 0 \\[1em] \Rightarrow k - 4 = 0 \text{ or } k - 8 = 0 \\[1em] \Rightarrow k = 4 \text{ or } k = 8 \\[1em]$

k ≠ 4 , as that will make a = (k - 4) = 0 and thus roots will become = ∞ .

Hence, the value of k is 8.