# How do you solve x^2-2x=5 by completing the square?

Jul 11, 2016

x = 1 ± sqrt(6)

#### Explanation:

Take the coefficient on the $x$-term, namely $- 2$, divide by $2$, and square the result, giving you

${\left(- \frac{2}{2}\right)}^{2} = {\left(- 1\right)}^{2} = 1$

Thus, we can now replace every ? mark in the expression

x^2-2x + ? = 5 + ?

with the number $1$, giving us

${x}^{2} - 2 x + 1 = 6$

We'd like two numbers whose product is $1$ and when added together gives us the result of $- 2$ (the number on the $x$-term).

Since $\left(- 1\right) \cdot \left(- 1\right) = 1$ and $\left(- 1\right) + \left(- 1\right) = - 2$, we now have our factors and can rewrite our expression in the following way:

$\left(x - 1\right) \left(x - 1\right) = 6$

or

${\left(x - 1\right)}^{2} = 6$

Taking the square root of both sides of the equation yields

sqrt((x-1)^2) = ± sqrt(6)

(x-1) = ± sqrt(6)

Adding $1$ to both sides gives us our final result of

x = 1 ± sqrt(6)