# How do you solve using completing the square method x^2+2x-3=0?

Apr 18, 2016

(x-1)(x+3)

#### Explanation:

There is no real method for this except just working it out in your head on which to brackets multiply to give ${x}^{2} + 2 x - 3$ except the quadratic formula which is good if they don't go into brackets easily $\frac{- b \sqrt{{b}^{2} - 4 \left(a\right) \left(c\right)}}{2 a}$ where a= ${x}^{2}$ b= $2 x$ and c=$3$

Apr 18, 2016

$x = 1 \textcolor{w h i t e}{\text{XX")orcolor(white)("XX}} x = - 3$
$\textcolor{w h i t e}{\text{XXX}}$(see below for "completing the square method" of solution)

#### Explanation:

Given ${x}^{2} + 2 x - 3 = 0$

Completing the Square
A squared binomial ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$

If ${x}^{2} + 2 x$ are the first two terms of such a squared binomial
then $a = 1$ (and ${a}^{2} = 1$)

We can add $\textcolor{red}{1}$ to complete the square but to keep the the equation correct we will need to subtract $\textcolor{red}{1}$ again.

${x}^{2} + 2 x \textcolor{red}{+ 1} - 3 \textcolor{red}{- 1} = 0$

${\left(x + 1\right)}^{2} - 4 = 0$

${\left(x + 1\right)}^{2} = 4$

$\left(x + 1\right) = \pm 2$

$x = - 1 + 2 = 1 \textcolor{w h i t e}{\text{XX")orcolor(white)("XX}} x = - 1 - 2 = - 3$