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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
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Solution
(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)`.
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