# Completing the Square

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Lesson Priority: High

- Learn how to solve a quadratic equation by completing the square
- Practice the simple case of this technique with quadratics that have a leading coefficient of 1
- Practice the general case of this technique with quadratics that have leading coefficients other than 1
- Know how to handle all steps of the process in cases when fraction coefficients are present in the initial problem

This technique for solving quadratics is often puzzling to students, since it involves some strange equation manipulation. But this technique has a few important uses in the future, and particularly is a topic that gets easier with practice!

Practice problems and worksheet coming soon!

## A Massive Manipulation Game

Completing the Square is a technique that many students are confused about at first, for two reasons. First, the actual technique looks like 100% smoke and mirrors tricks, since it entirely involves moving terms around in a clever way. Second, even after you see the steps and know how to proceed, it is not at all obvious why this manipulation technique is useful in any way. Read on to get closure on both of these issues, as well as see what kind of exam questions your teacher will throw at you.The main idea of completing the square is that we can take a quadratic equation and solve it using neither factoring methods nor the quadratic formula. The way we will proceed is to make one side of the equation a perfect square trinomial, and then take the square root of each side and solve for $x$.## Remembering Perfect Square Trinomials

First, let's recall what a perfect square trinomial even looks like. $x^2 - 6x + 9$ is a perfect square trinomial because$$(x-3)^2=x^2 -6x + 9$$Likewise, $x^2 + 5x - 7$ is not a perfect square trinomial, because there is no way to write this trinomial in the factored form $(x + a)^2$.In short, from what we learned in the past lesson on Perfect Square Trinomials », we know that we're dealing with a perfect square trinomial either if we start off with factored form that looks like $(x-a)^2$, or if we start off with standard form that we can factor into $(x-a)^2$ form. There is another property that standard form perfect square trinomials have, and it's key to the Completing the Square method.## Complete the Square - Expressions

The idea behind completing the square is that whatever $ax^2$ and $bx$ terms you have, you can figure out what the correct $c$ should be to make a perfect square. Add whatever constant to the expression that you need to achieve the correct $c$ value, and then subtract the same amount so that you effectively added $0$.Let's see an example to understand how it works.Example 1Complete the square of the following expression.$$x^2 + 20x +56$$$\blacktriangleright$ Because of the one-half squared rule we observed above for perfect square trinomials, we know that $x^2 + 20x$ can only be part of a perfect square if the constant term is one-half of $20$, then squared (which is $10^2$ or $100$). However, we don't have $100$ - we have $56$. But if we add $44$ and subtract $44$ simultaneously, we can group the pieces such that we have a perfect square trinomial.$$x^2 + 20x +56 + 44 - 44$$$$\longrightarrow \left(x^2 + 20x +56 + 44\right) - 44$$$$\longrightarrow (x + 10)^2 - 44$$There you have it - we have completed the square. By adding and subtracting the right number, we can rewrite the expression as a perfect square and an added or subtracted constant. It's all possible to do because of our ability to goal-seek the correct constant that we need using the "one-half $b$ then squared" rule of perfect square trinomials.## Complete the Square to Solve Equations

Now let's use the idea of creating perfect square trinomials to solve equations.Step 1 - Create a Perfect Square TrinomialThe quadratic we will be given in a problem will not be a perfect square - but we can make it a perfect square by adding and subtracting the right number on one side. Here is the first example we will work:$$x^2 -6x -2 = 5$$(1)Look at the left side. We want that to be a perfect square trinomial. We also know that if we have$$x^2 - 6x + \boxed{\, ? \,}$$That the missing number that would make it a perfect square integer is $9$, via the one-half middle squared rule. The typical way we will proceed is to simultaneously add and subtract that magic number on the left side.$$x^2 -6x - 2 + 9 - 9 = 5$$$$\longrightarrow \left( x^2 -6x + 9 \right) - 9 - 2 = 5$$(2)By adding and subtracting $9$ at the same time, we are in effect adding $0$ to the left side. Since we are adding $0$, we have not changed the equation. But now, we can group the perfect square trinomial as a separate item.## Mr. Math Makes It Mean

Most teachers will give you at least a few problems on a quiz or test where there is a constant other than $1$ on the quadratic term. We will be able to follow the same process, but there is just a little more work to do.Here's our example to learn by:Example 4$$3x^2 + 4x + 2 = \frac{102}{3}$$Step 1In order to create a perfect square trinomial using our $(b/2)^2$ trick, we need to work with a quadratic that has an $a$ coefficient of $1$. Since we do not have one, we have to factor out the leading coefficient so that we will have such an object to work with, even though the leading coefficient doesn't factor "nicely" out of the linear term.$$3x^2 + 4x + 2 = \frac{102}{3}$$$$\longrightarrow 3 \left(x^2 + \frac{4x}{3}\right) + 2 = \frac{102}{3}$$Now, when we create a perfect square trinomial, we will do so inside the parenthesis. Examining the quadratic inside the parenthesis (which is currently $x^2 + 4x/3$) we see that the $b$ term is $4/3$. Half of $4/3$ is $2/3$, and $2/3$ squared is $4/9$. We will add and subtract this amount, again, working inside the parenthesis.$$3 \left(x^2 + \frac{4x}{3} + \frac{4}{9} -\frac{4}{9} \right) + 2 = \frac{102}{3}$$Step 2Now we'll factor the perfect square trinomial we created. We are still keeping it inside the parenthesis for now.$$3 \left(\left(x+ \frac{2}{3}\right)^2 - \frac{4}{9} \right) + 2 = \frac{102}{3}$$Step 3We must isolate the $(x + 2/3)^2$ term. This is the part you need to pay attention to, because students rush at this point, when instead they need to be careful. The term $-\frac{4}{9}$ in the parenthesis comes out of the parenthesis as $-\frac{4}{3}\,\,$, because it needs to be multiplied by the front coefficient of $3$. A quick look at this symbolically to help say this a different way:$$3(a + b + c + d) = 3(a + b + c) + 3d$$Now our equation looks like$$3 \left(x+ \frac{2}{3}\right)^2 - \frac{4}{3} + 2 = \frac{102}{3}$$And, cleaning up the constants$$3 \left(x+ \frac{2}{3}\right)^2 + \frac{2}{3} = \frac{102}{3}$$$$\rightarrow 3 \left(x+ \frac{2}{3}\right)^2 = \frac{100}{3}$$$$\rightarrow \frac{\cancel{3} \left(x+ \frac{2}{3}\right)^2}{\cancel{3}} = \frac{\frac{100}{3}}{3}$$$$\rightarrow \left(x+ \frac{2}{3}\right)^2 = \frac{100}{9}$$Step 4Now let's take the square root of both sides.$$\sqrt{\left(x + \frac{2}{3}\right)^2} = \sqrt{\frac{100}{9}}$$$$x + \frac{2}{3} = \pm \frac{10}{3}$$Step 5Move the $2/3$ over to the left side to isolate $x$ and finish the problem.$$x = -\frac{2}{3} \pm \frac{10}{3}$$$$\boxed{\Rightarrow x = -4, \,\, \frac{8}{3}}$$## Put It To The Test

Because this process is repetitive and structured, a little practice will set you straight really quickly for a quiz or test. Make sure you can do the problems that require square root simplification and / or working with fractions, as teachers tend to universally give a few on tests to sift out the top scoring students.Examples 5 to 7: rewrite each expression as the sum of a squared binomial and a constant, using the Completing the Square method.- Understand what kind of equations and situations we use the Complete the Square method for
- Recall what perfect square trinomials are and be able to manipulate a quadratic into becoming one
- Be able to execute the Complete the Square method for basic cases when $a$ is 1
- Master the method for more involved cases, such as fraction coefficients and the $a$ coefficient being different than $1$
- Turn in answers that are simplified sufficiently, including a single fraction answer when applicable, or simplified square roots when applicable

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