# How do you solve using the completing the square method x^2 + 2x - 63 = 0?

Mar 8, 2016

$\left(x + 9\right) \left(x - 7\right)$

#### Explanation:

To solve the equation x^2+2x−63=0, using the completing the square method, one has to complete the square using the variable $x$.

We know the identity ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$, hence we should halve the coefficient of $x$ and square it and then add and subtract the 'square' in the trinomial. Here coefficient of $x$ is 2 and hence squareof half of it is $1$. Hence

x^2+2x−63=0 can be written as

x^2+2x+1-1−63=0 or

$\left({x}^{2} + 2 \times x \times 1 + {1}^{2}\right) - 64$ or

${\left(x + 1\right)}^{2} - {8}^{2}$

Now using the identity ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$, this becomes

$\left(\left(x + 1\right) + 8\right) \times \left(\left(x + 1\right) - 8\right)$ or

$\left(x + 9\right) \left(x - 7\right)$