Question
Without solving the following quadratic equation, find the value of 'm' for which the given equation has real and equal roots.
$x^2+ 2 (m - 1) x + (m + 5) = 0.$
$x^2+ 2 (m - 1) x + (m + 5) = 0.$
✓
Answer
$x^2+ 2 (m - 1) x + (m + 5) = 0$
Equating with $ax^2 + bx + c = 0$
$a = 1, b = 2 (m - 1), c = (m + 5)$
Since equation has real and equal roots.
so, $D = 0$
$\Rightarrow b^2 - 4ac = 0$
$\Rightarrow [2 (m - 1)^2 - 4 x 1 x (m + 5) = 0$
$\Rightarrow 4 (m - 1)^2 - 4 (m + 5) = 0$
$\Rightarrow 4 [m^2- 2m + 1 - m - 5)] = 0$
$\Rightarrow m^2 - 3m - 4 = 0$
$\Rightarrow (m + 1) (m - 4) = 0$
Either $m + 1 = 0$
$m = -1$
or
$m - 4 = 0$
$m = 4$
$m = -1, 4$
Equating with $ax^2 + bx + c = 0$
$a = 1, b = 2 (m - 1), c = (m + 5)$
Since equation has real and equal roots.
so, $D = 0$
$\Rightarrow b^2 - 4ac = 0$
$\Rightarrow [2 (m - 1)^2 - 4 x 1 x (m + 5) = 0$
$\Rightarrow 4 (m - 1)^2 - 4 (m + 5) = 0$
$\Rightarrow 4 [m^2- 2m + 1 - m - 5)] = 0$
$\Rightarrow m^2 - 3m - 4 = 0$
$\Rightarrow (m + 1) (m - 4) = 0$
Either $m + 1 = 0$
$m = -1$
or
$m - 4 = 0$
$m = 4$
$m = -1, 4$
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