# Completing the Square

## Key Questions

• Since factoring is usually quicker, I would always try factoring first; however, factoring is not always easy, so if that is the case, then I would try completing the square or the quadratic formula next.

I hope that this was helpful.

• Can every quadratic be solved by using the completing the square method? Yes, it sure seems so.....but, I wouldn't want to do it that way every time!!

Certainly, every quadratic can be solved by the quadratic formula, but I also wouldn't want to use it every time either.

I think you should develop some strategies for which method is best in which circumstances. Purple Math link

There are even websites that will solve the problems for you: Solver

For completing the square, I look for two things: "1" as the lead coefficient (on the ${x}^{2}$, and an even number for the coefficient on the linear term (the x term).

Example: Solve ${x}^{2} + 4 x - 7 = 0$
Step 1: ${x}^{2} + 4 x = 7$ (move the constant to the opposite side)
Step 2: take half of the "4", and square that number. ${2}^{2} = 4$
Step 3: Add that number to both sides ${x}^{2} + 4 x + 4 = 7 + 4$
Step 4: Factor the trinomial: ${\left(x + 2\right)}^{2} = 11$
Step 5: Take the square root of both sides: $\sqrt{{\left(x + 2\right)}^{2}} = \sqrt{11}$
Step 6: $x + 2 = \pm \sqrt{11}$ be sure to use the $\pm$ on the square root!
Step 7: $x = - 2 \pm \sqrt{11}$ move the constant to the other side.

Phew, that's a lot of steps! And it definitely takes practice. One more example:

Solve ${x}^{2} - 8 x - 9 = 0$
${x}^{2} - 8 x = 9$
${x}^{2} - 8 x + 16 = 9 + 16$
${x}^{2} - 8 x + 16 = 25$
${\left(x - 4\right)}^{2} = 25$
$\sqrt{{\left(x - 4\right)}^{2}} = \sqrt{25}$
$x - 4 = \pm 5$
$x = 4 + 5 \mathmr{and} 4 - 5$
so, x = 9 or -1. Wow!

Of course, those nice, rational answers tell me that the original problem could have been solved by factoring: (x-9)(x+1)=0
Use the zero product property to solve x = 9 or x = -1.

See explanation below

#### Explanation:

When you have a polynomial such as

${x}^{2} + 4 x + 20$

it is sometimes desirable to express it in the form of

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

To do this, we can artificially introduce a constant which allows us to factor a perfect square out of the expression like so:

${x}^{2} + 4 x + 20$

$= {x}^{2} + 4 x + \textcolor{red}{4} - \textcolor{g r e e n}{4} + 20$

Notice that by simultaneously adding and subtracting $4$, we have not changed the value of the expression.

Now we can do this:

$= \left({x}^{2} + 4 x + \textcolor{red}{4}\right) + \left(20 - \textcolor{g r e e n}{4}\right)$

$= {\left(x + 2\right)}^{2} + 16$

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

We have "completed the square"!

To simplify quadratic expressions so that they become solvable with square roots.

#### Explanation:

Completing the square is an example of a Tschirnhaus transformation - the use of a substitution (albeit implicitly) in order to reduce a polynomial equation to simpler form.

So given:

$a {x}^{2} + b x + c = 0 \text{ }$ with $a \ne 0$

we could write:

$0 = 4 a \left(a {x}^{2} + b x + c\right)$

$\textcolor{w h i t e}{0} = 4 {a}^{2} {x}^{2} + 4 a b x + 4 a c$

$\textcolor{w h i t e}{0} = {\left(2 a x\right)}^{2} + 2 \left(2 a x\right) b + {b}^{2} - \left({b}^{2} - 4 a c\right)$

$\textcolor{w h i t e}{0} = {\left(2 a x + b\right)}^{2} - {\left(\sqrt{{b}^{2} - 4 a c}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(2 a x + b\right) - \sqrt{{b}^{2} - 4 a c}\right) \left(\left(2 a x + b\right) + \sqrt{{b}^{2} - 4 a c}\right)$

$\textcolor{w h i t e}{0} = \left(2 a x + b - \sqrt{{b}^{2} - 4 a c}\right) \left(2 a x + b + \sqrt{{b}^{2} - 4 a c}\right)$

Hence:

$2 a x = - b \pm \sqrt{{b}^{2} - 4 a c}$

So:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

So having started with a quadratic equation in the form:

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

we got it into a form ${t}^{2} - {k}^{2} = 0$ with $t = \left(2 a x + b\right)$ and $k = \sqrt{{b}^{2} - 4 a c}$, eliminating the linear term leaving only squared terms.

So long as we are happy calculating square roots, we can now solve any quadratic equation.

Completing the square is also useful for getting the equation of a circle, ellipse or other conic section into standard form.

For example, given:

${x}^{2} + {y}^{2} - 4 x + 6 y - 12 = 0$

completing the square we find:

${\left(x - 2\right)}^{2} + {\left(y + 3\right)}^{2} = {5}^{2}$

allowing us to identify this equation as that of a circle with centre $\left(2 , - 3\right)$ and radius $5$.

• (This is going to take a minute or two.)

Completing incomplete squares:

Background:
The square of an expression of the form $x + n$ or $x - n$ is:
${\left(x + n\right)}^{2} = {x}^{2} + 2 n x + {n}^{2}$ or
${\left(x - n\right)}^{2} = {x}^{2} - 2 n x + {n}^{2}$.
Notice: the sign on the middle term matches the sign in the middle of the binomial on the left AND the last term is positive in both.
Also notice that if we allow $n$ to be negative, we only need to write and think about ${\left(x + n\right)}^{2} = {x}^{2} + 2 n x + {n}^{2}$ (The sign in the midde will match the sign of $n$.)

Completing the square:
An expression like ${x}^{2} + 6 x$ may be thought of as an "incomplete" square. To make it complete, we'd need to add ${n}^{2}$ to the expression. We can figure out what to use for $n$ by realizing that the $6 x$ in the middle need to be $2 n x$. So, we want $n = 3$ and ${n}^{2} = 9$.

Of course you can's just add a number to an expression without changing the value of the expression, so if we want to keep the same value we'll have to make up for adding $9$.
We do this by also subtracting $9$. That doesn't change the value of ${x}^{2} + 6 x$, but it does change the way it's written.

We write ${x}^{2} + 6 x + 9 - 9$ If we group it this way: $\left({x}^{2} + 6 x + 9\right) - 9$ then we have a perfect square minus $9$

And $\left({x}^{2} + 6 x + 9\right) - 9 = {\left(x + 3\right)}^{2} - 9$

Solving an equation by completing the square:

Solve: ${x}^{2} + 6 x - 16 = 0$ (by completing the square)

Each of the following equations is equivalent (has exactly the same solutions) as the lines before it.

${x}^{2} + 6 x - 16 = 0$
${x}^{2} + 6 x = 16$
${x}^{2} + 6 x + 9 - 9 = 16$.
${x}^{2} + 6 x + 9 = 16 + 9$. So the first equation is equivalent to
${\left(x + 3\right)}^{2} = 25$

And the last equation above is satisfied exactly when:
$x + 3 = - 5$ or $x + 3 = 5$. So $x = - 8$ or $x = 2$.

The solution set to the first equation is: $\left\{- 8 , 2\right\}$.

I'll post another (more challenging) example too.