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

(i) (3m + 1)x² + 2(m + 1)x + m = 0

(ii) x² + 2(m - 1)x + (m + 5) = 0

(i) The given equation is (3m + 1)x² + 2(m + 1)x + m = 0.

Comparing with ax² + bx + c = we obtain,
a = 3m + 1 , b = 2(m + 1) , c = m

$\therefore \text{Discriminant} = b^2 - 4ac \\[0.5em] = (2m + 2)^2 - 4 \times 3m + 1 \times m \\[0.5em] = 4m^2 + 4 + 8m - 4m(3m + 1) \\[0.5em] = 4m^2 + 4 + 8m - 12m^2 - 4m \\[0.5em] = 4m^2 - 12m^2 + 8m - 4m + 4 \\[0.5em] = -8m^2 + 4m + 4 \\[0.5em]$

For equal roots, discriminant = 0

$\Rightarrow -8m^2 + 4m + 4 = 0 \\[0.5em] \Rightarrow -4(2m^2 - m - 1) = 0 \\[0.5em] \Rightarrow 2m^2 - m - 1 = 0 \\[0.5em] \Rightarrow 2m^2 - 2m + m - 1 = 0 \\[0.5em] \Rightarrow 2m(m - 1) + 1(m - 1) = 0 \\[0.5em] \Rightarrow (2m + 1)(m - 1) = 0 \\[0.5em] \Rightarrow 2m + 1 = 0 \text{ or } m - 1 = 0 \\[0.5em] \Rightarrow m = -\dfrac{1}{2} \text{ or } m = 1$

Hence, the value of m is $-\dfrac{1}{2}$ and 1.

(ii) The given equation is x² + 2(m - 1)x + (m + 5) = 0.

Comparing with ax² + bx + c = we obtain,
a = 1 , b = 2(m - 1) , c = m + 5

$\therefore \text{Discriminant} = b^2 - 4ac \\[0.5em] = (2m - 2)^2 - 4 \times 1 \times m + 5 \\[0.5em] = 4m^2 + 4 - 8m - 4(m + 5) \\[0.5em] = 4m^2 + 4 - 8m - 4m - 20 \\[0.5em] = 4m^2 - 12m - 16 \\[0.5em] = 4m^2 - 12m - 16 \\[0.5em]$

For equal roots, discriminant = 0

$\Rightarrow 4m^2 - 12m - 16 = 0 \\[0.5em] \Rightarrow 4(m^2 - 3m - 4) = 0 \\[0.5em] \Rightarrow m^2 - 3m - 4 = 0 \\[0.5em] \Rightarrow m^2 - 4m + m - 4 = 0 \\[0.5em] \Rightarrow m(m - 4) + 1(m - 4) = 0 \\[0.5em] \Rightarrow (m - 4)(m + 1) = 0 \\[0.5em] \Rightarrow m - 4 = 0 \text{ or } m + 1 = 0 \\[0.5em] \Rightarrow m = 4 \text{ or } m = -1$

Hence, the value of m is 4, -1.