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Question
Without solving the following quadratic equation, find the value of 'm' for which the given equation has real and equal roots.
x2 + 2(m – 1)x + (m + 5) = 0
Solution
x2 + 2(m – 1)x + (m + 5) = 0
Equating with ax2 + bx + c = 0
a = 1, b = 2(m – 1), c = (m + 5)
Since equation has real and equal roots.
So, D = 0
`=>` b2 – 4ac = 0
`=>` [2(m – 1)2 – 4 × 1 × (m + 5) = 0
`=>` 4(m – 1)2 – 4 (m + 5) = 0
`=>` 4[m2 – 2m + 1 – m – 5)] = 0
`=>` m2 – 3m – 4 = 0
`=>` (m + 1)(m – 4) = 0
Either m + 1 = 0 or m – 4 = 0
`=>` m = –1 or m = 4
Hence, the values of m are –1, 4.
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Solution :
Compare x2 + 2x – 9 = 0 with ax2 + bx + c = 0
a = 1, b = 2, c = `square`
∴ b2 – 4ac = (2)2 – 4 × `square` × `square`
Δ = 4 + `square` = 40
∴ b2 – 4ac > 0
∴ The roots of the equation are real and unequal.
