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

(i) 9x² + kx + 1 = 0

(ii) x² - 2kx + 7k - 12 = 0 Also, find the roots for those values of k in each case.

(i) The given equation is 9x² + kx + 1 = 0.

Comparing with ax² + bx + c = we obtain,
a = 9 , b = k , c = 1

$\therefore \text{Discriminant} = b^2 - 4ac \\[0.5em] = (k)^2 - 4 \times 9 \times 1 \\[0.5em] = k^2 - 36 \\[0.5em]$

For equal roots, discriminant = 0

$\Rightarrow k^2 - 36 = 0 \\[0.5em] \Rightarrow k^2 - (6)^2 = 0 \\[0.5em] \Rightarrow (k + 6)(k - 6) = 0 \\[0.5em] \Rightarrow k + 6 = 0 \text{ or } k - 6 = 0 \\[0.5em] \Rightarrow k = -6 \text{ or } k = 6 \\[0.5em]$

When k = 6 the equation becomes 9x² + 6x + 1 = 0 so the roots are :

The roots of the equation are given by

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-(6) ± \sqrt{(6)^2 - 4\times 9 \times 1}}{2 \times 9} \\[1em] \Rightarrow x = \dfrac{-6 ± \sqrt{36 - 36}}{18} \\[1em] \Rightarrow x = \dfrac{-6 ± \sqrt{0}}{18} \\[1em] \Rightarrow x = \dfrac{-6 +0}{18} \text{ or } \dfrac{-6 - 0}{18} \\[1em] \Rightarrow x = -\dfrac{1}{3} \text{ or } -\dfrac{1}{3} \\[1em]$

When k = -6 the equation becomes 9x² - 6x + 1 = 0 so the roots are :

The roots of the equation are given by

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-(-6) ± \sqrt{(-6)^2 - 4\times 9 \times 1}}{2 \times 9} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{36 - 36}}{18} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{0}}{18} \\[1em] \Rightarrow x = \dfrac{6 +0}{18} \text{ or } \dfrac{6 - 0}{18} \\[1em] \Rightarrow x = \dfrac{1}{3} \text{ or } \dfrac{1}{3} \\[1em]$

Hence, the values of k are 6, -6 ; when k = 6, roots are $-\dfrac{1}{3}, -\dfrac{1}{3}$ and when k = -6, roots are $\dfrac{1}{3}, \dfrac{1}{3}$.

(ii) The given equation is x² - 2kx + 7k - 12 = 0.

Comparing with ax² + bx + c = we obtain,
a = 1 , b = -2k , c = 7k - 12

$\therefore \text{Discriminant} = b^2 - 4ac \\[0.5em] = (-2k)^2 - 4 \times 1 \times (7k - 12) \\[0.5em] = 4k^2 - 4(7k - 12) \\[0.5em] = 4k^2 - 28k + 48$

For equal roots, discriminant = 0

$\Rightarrow 4k^2 - 28k + 48 = 0 \\[0.5em] \Rightarrow 4(k^2 - 7k + 12) = 0 \\[0.5em] \Rightarrow k^2 - 7k + 12 = 0 \\[0.5em] \Rightarrow k^2 - 4k - 3k + 12 = 0 \\[0.5em] \Rightarrow k(k - 4) - 3(k - 4) = 0 \\[0.5em] \Rightarrow (k - 3)(k - 4) = 0 \\[0.5em] \Rightarrow k - 3 = 0 \text{ or } k - 4 = 0 \\[0.5em] \Rightarrow k = 3 \text{ or } k = 4 \\[0.5em]$

When k = 3 the equation becomes x² - 6x + 9 = 0 so the roots are :

The roots of the equation are given by

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-(-6) ± \sqrt{(-6)^2 - 4\times 1 \times 9}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{36 - 36}}{2} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{0}}{12} \\[1em] \Rightarrow x = \dfrac{6 +0}{2} \text{ or } \dfrac{6 - 0}{2} \\[1em] \Rightarrow x = 3 \text{ or } 3 \\[1em]$

When k = 4 the equation becomes x² - 8x + 16 = 0 so the roots are :

The roots of the equation are given by

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-(-8) ± \sqrt{(-8)^2 - 4\times 1 \times 16}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{8 ± \sqrt{64 - 64}}{2} \\[1em] \Rightarrow x = \dfrac{8 ± \sqrt{0}}{2} \\[1em] \Rightarrow x = \dfrac{8 + 0}{2} \text{ or } \dfrac{8 - 0}{2} \\[1em] \Rightarrow x = 4 \text{ or } 4 \\[1em]$

Hence the values of k are 3, 4 ; when k = 3, roots are 3, 3 and when k = 4, roots are 4, 4.