# How do you solve x^2 - 5x + 1 = 0 by completing the square?

Jul 28, 2015

#### Answer:

$\left\{\begin{matrix}{x}_{1} = \frac{5}{2} + \frac{\sqrt{21}}{2} \\ {x}_{2} = \frac{5}{2} - \frac{\sqrt{21}}{2}\end{matrix}\right.$

#### Explanation:

Starting from the general form of the quadratic equation

$a {x}^{2} + b x + c = 0$

get your quadratic to the form

${x}^{2} + \frac{b}{a} x = - \frac{c}{a}$

In your case, $a = 1$, which means that all you have to do is move $1$ on the other side of the equation

${x}^{2} - 5 x + \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} = 0 - 1$

${x}^{2} - 5 x = - 1$

Now, in order to solve this equation by completing the square, you need to find a term to add to both sides of the equation such that the left side of the equation can be written as the square of a binomial.

The coefficient of the $x$-term will help you determine what term you need to add to both sides of the equation.

So, divide this coefficient by 2, then square the result. In your case, you have

$\frac{\left(- 5\right)}{2} = - \frac{5}{2}$, then

${\left(- \frac{5}{2}\right)}^{2} = \frac{25}{4}$

This means that the quadratic becomes

${x}^{2} - 5 x + \frac{25}{4} = - 1 + \frac{25}{4}$

Using the formula for the square of a binomial

$\textcolor{b l u e}{{\left(x - n\right)}^{2} = {x}^{2} - 2 n + {n}^{2}}$

the left side of the equation can now be written as

${x}^{2} - 5 x + \frac{25}{4} = {x}^{2} - 2 \left(\frac{5}{2}\right) x + {\left(- \frac{5}{2}\right)}^{2} = {\left(x - \frac{5}{2}\right)}^{2}$

Your quadratic is now equivalent to

${\left(x - \frac{5}{2}\right)}^{2} = \frac{21}{4}$

Take the square root of both sides to get

$x - \frac{5}{2} = \pm \sqrt{\frac{21}{4}} = \pm \frac{\sqrt{21}}{2}$

The two solutions to this equation will be

${x}_{1} = \textcolor{g r e e n}{\frac{5}{2} + \frac{\sqrt{21}}{2}}$ and ${x}_{2} = \textcolor{g r e e n}{\frac{5}{2} - \frac{\sqrt{21}}{2}}$