# How do you solve using the completing the square method x^2 - 5x = 9?

Jan 20, 2017

Take the coefficient of the $x$ term ($- 5$) and:
1. Divide it by 2 (to get $- \frac{5}{2}$).
2. Square this (to get $\frac{25}{4}$).
3. Add this final value to both sides.

#### Explanation:

Completing the square means seeking a constant term $n$ to add to ${x}^{2} - 5 x$, so that ${x}^{2} - 5 x + n$ is a perfect square.

First, let's look at what happens when we FOIL a perfect square binomial of the form ${\left(x + a\right)}^{2}$:

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

A perfect square will always have a distributed form like this.

What we notice is that if we take the coefficient of the $x$ term ($2 a$), cut it in half, and then square it, we get ${a}^{2}$, the constant term. Thus, if given ${x}^{2} + 2 a x = b$, we would complete the square by adding ${a}^{2}$ (that is, the square of half of $2 a$) to both sides, giving

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

so that the trinomial on the left will be guaranteed to be a perfect square—the square of $\left(x + a\right)$.

For this particular problem, we are given ${x}^{2} - 5 x = 9$. So, $- 5$ is like our "$2 a$". And if

$- 5 = 2 a$,

then

$a = - \frac{5}{2}$,

and

${a}^{2} = \frac{25}{4}$.

Thus, ${x}^{2} - 5 x + \frac{25}{4}$ is the completed square we seek, meaning we need to add $\frac{25}{4}$ to both sides:

${x}^{2} - 5 x + \frac{25}{4} = 9 + \frac{25}{4}$

Okay, so if this ${x}^{2} - 5 x + \frac{25}{4}$ is a perfect square, what is its factored form? (Or, what is its square root?)

That's easy—remember that the factored form of ${x}^{2} + \textcolor{red}{2 a} x + {a}^{2}$ is ${\left(x + \textcolor{b l u e}{a}\right)}^{2}$. The $\textcolor{b l u e}{a}$ in the factor is half of the $\textcolor{red}{2 a}$ in the trinomial. So the factored form of ${x}^{2} - 5 x + \frac{25}{4}$ will be ${\left(x - \frac{5}{2}\right)}^{2}$, because $- \frac{5}{2}$ is half of $- 5$.

We simplify both sides now to get

${\left(x - \frac{5}{2}\right)}^{2} = \frac{61}{4}$

Now our LHS is a perfect square, so we can solve for $x$ by taking the square root of both sides:

$x - \frac{5}{2} = \pm \frac{\sqrt{61}}{2}$

and then adding $\frac{5}{2}$ to both sides:

$x = \frac{5}{2} \pm \frac{\sqrt{61}}{2} \text{ "=" } \frac{5 \pm \sqrt{61}}{2}$.

## Note:

If the coefficient on the ${x}^{2}$ term is something other than $1$, you'll want to either factor it out or divide everything by it first, so that this method will work.