How do you find the roots for x^2 + 2x – 48 = 0 ?

May 16, 2015

Let us try completing the square

$0 = {x}^{2} + 2 x - 48$

$= {\left(x + 1\right)}^{2} - 1 - 48 = {\left(x + 1\right)}^{2} - 49 = {\left(x + 1\right)}^{2} - {7}^{2}$

So ${\left(x + 1\right)}^{2} = 7$

So $x + 1 = \pm \sqrt{7}$

So $x = - 1 \pm \sqrt{7}$

In general

$a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$