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Question
Without solving the following quadratic equation, find the value of m for which the given equation has equation has real and equal roots.
`x^2 + 2(m - 1)x + (m + 5) = 0`
Solution
Given Quadratic Equation is
`x^2 + 2(m - 1)x + (m + 5) = 0`
Here a = 1,b = 2(m - 1) and c = (m + 5)
Discriminant isgiven by `D^2 = b^2 - 4ac`
For Real and equal roots, D = 0
`=> b^2 - 4ac = 0`
`=> [2(m - 1)]^2 - 4(m + 5) = 0`
`=> m^2 + 1 - 2m - m - 5 = 0`
`=> m^2 - 3m - 4 = 0`
Factorising we get (m + 1)(m - 4) = 0
=> m = -1 or m = 4
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