Without solving the following quadratic equation, find the value of $m$ for which the given equation has real and equal roots: $\;$ $x^2 + 2 \left(m - 1\right) x + \left(m + 5\right) = 0$
Given quadratic equation: $\;$ $x^2 + 2 \left(m - 1\right) x + \left(m + 5\right) = 0$
Comparing with the standard quadratic equation $\;$ $ax^2 + bx + c = 0$ $\;$ gives
$a = 1, \; b = 2\left(m - 1\right), \; c = m + 5$
The given quadratic equation has real and equal roots when its discriminant $= \Delta = b^2 - 4ac = 0$
$\therefore \;$ We have
$\Delta = 4 \left(m - 1\right)^2 - 4 \times 1 \times \left(m + 5\right) = 0$
i.e. $\;$ $m^2 -2m + 1 = m + 5$
i.e. $\;$ $m^2 - 3m - 4 = 0$
i.e. $\;$ $m^2 -4m + m - 4 = 0$
i.e. $\;$ $m \left(m - 4\right) + 1 \left(m - 4\right) = 0$
i.e. $\left(m + 1\right) \left(m - 4\right) = 0$
i.e. $\;$ $m + 1 = 0$ $\;$ OR $\;$ $m - 4 = 0$
$\implies$ $m = -1$ $\;$ OR $\;$ $m = 4$