Question

Find the value(s) of m for which each of the following quadratic equation has real and equal roots: (3m + 1)x² + 2(m + 1)x + m = 0

Answer 1

(3m + 1)x² + 2(m + 1)x + m = 0
Here a = 3m + 1, b = 2(m + 1), c = m
D = b² - 4ac
= [2(m + 1)]² - 4 x (3m + 1)(m)
= 4(m² + 2m + 1) - 12m² - 4m
= 4m² + 8m + 4 - 12m² - 4m
= -8m² + 4m + 4
∴ Roots are equal.
∴ D = 0
⇒ -8m² + 4m + 4 = 0
⇒ 2m² - m - 1 = 0    ...(Dividing by 4)
⇒ 2m² - 2m + m - 1 = 0
⇒ 2m(m - 1) + 1(m - 1) = 0
⇒ (m - 1)(2m + 1) = 0
Either m - 1 = 0,
then m = 1
or
2m + 1 = 0,
then 2m = -1

⇒ m = `-(1)/(2)`.