Find the value(s) of k for which each of the following quadratic equation has equal roots:

(i) 3kx² = 4(kx - 1)

(ii) (k + 4)x² + (k + 1)x + 1 = 0

Also, find the roots for that value(s) of k in each case.

(i) Given,

$3kx^2 = 4(kx - 1) \\[0.5em] \Rightarrow 3kx^2 - 4(kx - 1) = 0 \\[0.5em] \Rightarrow 3kx^2 - 4kx + 4 = 0 \\[0.5em]$

Comparing it with ax² + bx + c = 0
a= 3k, b = -4k, c = 4

Given,

Equation has equal roots

∴ b² - 4ac = 0

$\Rightarrow (-4k)^2 - 4 \times 3k \times 4 = 0 \\[0.5em] \Rightarrow 16k^2 - 48k = 0 \\[0.5em] \Rightarrow 16k(k - 3) = 0 \\[0.5em] \Rightarrow 16k = 0 \text{ or } k - 3 = 0 \\[0.5em] k = 0 \text{ or } k = 3$

But k cannot be equal to 0 as that will make a = 3k = 0 , which will make roots equal to infinity.

∴ k = 3

Hence equation is 9x² - 12x + 4 = 0

$\Rightarrow 9x^2 - 6x - 6x + 4 = 0 \\[0.5em] \Rightarrow 3x(3x - 2) - 2(3x - 2) = 0 \\[0.5em] \Rightarrow (3x - 2)(3x - 2) = 0 \\[0.5em] \Rightarrow 3x - 2 = 0 \text{ or } 3x - 2 = 0 \\[0.5em] x = \dfrac{2}{3} \text{ or } x = \dfrac{2}{3}$

Hence the value of k is 3 and the roots are $\dfrac{2}{3}, \dfrac{2}{3}.$

(ii) The given equation is (k + 4)x² +(k + 1)x + 1 = 0

Comparing it with ax² + bx + c = 0
a = (k + 4), b = (k + 1), c = 1

Given,

Equation has equal roots

∴ b² - 4ac = 0

$\Rightarrow (k + 1)^2 - 4 \times (k + 4) \times 1 = 0 \\[1em] \Rightarrow (k^2 + 1 + 2k) - 4(k + 4) = 0 \\[1em] \Rightarrow k^2 + 1 + 2k - 4k - 16 = 0 \\[1em] \Rightarrow k^2 - 2k - 15 = 0 \\[1em] \Rightarrow k^2 - 5k + 3k - 15 = 0 \\[1em] \Rightarrow k(k - 5) + 3(k - 5) = 0 \\[1em] \Rightarrow (k + 3)(k - 5) = 0 \\[1em] \Rightarrow k + 3 = 0 \text{ or } k - 5 = 0 \\[1em] k = -3 \text{ or } k = 5$

∴ When k = -3 , equation is x² - 2x + 1 = 0

$\Rightarrow x^2 - x - x + 1 = 0 \\[1em] \Rightarrow x(x - 1) - 1(x - 1) = 0 \\[1em] \Rightarrow (x - 1)(x - 1) = 0 \\[1em] \Rightarrow x - 1 = 0 \text{ or } x - 1 = 0 \\[1em] x = 1 \text{ or } x = 1$

∴ When k = 5 , equation is 9x² + 6x + 1 = 0

$\Rightarrow 9x^2 + 3x + 3x + 1 = 0 \\[1em] \Rightarrow 3x(3x + 1) + 1(3x + 1) = 0 \\[1em] \Rightarrow (3x + 1)(3x + 1) = 0 \\[1em] \Rightarrow 3x + 1 = 0 \text{ or } 3x + 1 = 0 \\[1em] x = -\dfrac{1}{3} \text{ or } -\dfrac{1}{3}$

k = -3, 5 When k = -3 , roots are 1, 1 When k = 5 , roots are $-\dfrac{1}{3}, -\dfrac{1}{3}$