## Completing the Square to Solve

In the previous two sections we saw you can change a quadratic in standard form into vertex form by completing the square. In this section we will see how you can solve a quadratic equation, set equal to zero, by using completing the square.

The idea shared here is the same as the previous two sections, but the steps are slightly different because we have a different application. This skill and concept, of completing the square, is highly adaptable and useful in many areas of mathematics. This is another opportunity to “really get it,” so be patient with yourself! Give yourself time, ask questions, be patient, you’ll get it.

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This is another application of Completing the Square (CTS), which is a very difficult skill to learn in future mathematics. Right now is a great time to learn this process and concept as it is fairly isolated, without too many other issues surrounding it. There are other ways of solving quadratic equations that are not factorable, which we will learn. But this is a great application and will make a lot of things, like the quadratic formula, make sense.

Let’s get to it. Suppose you’re trying to find the *x* – intercepts of a quadratic equation that was NOT factorable. With what you’ve learned so far, that would be impossible. You need some more tools at your disposal. We can change the equation into vertex form, and then use inverse operations. Let’s see how that looks.

Solve the equation by Completing the Square:

We will use the same concepts as we did early, when changing from Standard Form to Vertex Form. But, because our end goal is slightly different, we will change a few things around to make the process more efficient. Let’s first move *c* to the other side, by using inverse operations.

Notice how we left space after the *bx* term. This is where we will add a new value to make the left side a perfect square. To do so, we must take half of *b*, and square it. This will be our new *c* value. But, if we add it to the left, we must also either subtract it from the left, *OR* add it also to the right side of the equal sign.

Remember, on the left we do NOT want to square the number because this is our factor. On the right side, we do because we’ll combine like terms there.

Now, factor the left, simplify the right.

Remember, do not use decimals. Fractions work out easy. Just reduce and follow the order of operations.

Now, let’s use inverse operations to solve for *x*. If there was an *a* term, we’d divide both sides by *a*, but it is 1, so we can skip that. We will have to take the square root of both sides. Note that the square root of 4 is 2. Also, remember, square roots have two answers, a positive and a negative answer.

See how easy the fractions work out. They are already like denominators. Just add 3/2 to both sides, set up your two equations, one plus and one subtraction.

These are **exact solutions.** The square root of 13 is irrational, meaning to write it as a decimal would be to approximate the number. We could do so, and will need to if we were being asked to graph.

Since 13 is between 9 and 16, the square root of 13 will be between 9 and 16. It is close to 3.6, which we will use as an approximate.

Our *x* – intercepts would be (3.3, 0) and (-0.3, 0).

The process is slightly different when the leading coefficient is NOT 1, as in *y* = 2*x*^{2} – 6*x* – 1. However, you handle the *a* coefficient just as you did when changing to Vertex Form from Standard Form. Let’s see how that looks.

Step one: Move *c* to the other side using inverse operations. At the same time, divide all terms by *a*.

Complete the square, keeping the equation balanced by adding the same value to each side.

Now, factor the left and combine like terms on the right.

Now, use inverse operations to solve for *x*. Be sure to account for both the positive and negative roots.

The exact *x* – intercepts are To figure out the approximations you would need to find a suitable decimal approximation for the square root of 11, then perform the arithmetic.

Summary: The process of solving by completing the square works because we take a non-square quadratic and change it into a perfect square, algebraically. Then, we use inverse operations to solve for *x*. The process will always work, but is tedious. Finding the *x* – intercepts by factoring is superior as it is less complicated, quicker, and has fewer potential pitfalls. However, not all quadratic equations are factorable.

Coming Soon