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17calculus > precalculus > completing the square

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Completing The Square

The idea of completing the square is just to move the terms around so that an expression is in the form we want, without changing the problem. What we want to end up with is a term that is squared (thus the name completing the square). We may also end up with an extra constant. To do this, we need some basic guidelines.

Basic Guidelines For Completing The Square

guideline 1 - - We may add only zero, i.e. if we add a number somewhere, we are required to also subtract the same number.
Now, this can get confusing if you are not paying attention since there are actually two way to do this. First, is to add and subtract the same number on the same side of the equal sign. For example, in \(2x=7 \to 2x+a-a=7\), we added and subtracted a on the left of the equal sign. This is the best way to do it and, in our opinion, the most understandable. When you are working problems and, especially when you are first learning this technique, it is always best to think about it this way.

This second way is equivalent but looks a bit different. Instead of adding and subtracting the same number on one side of the equal sign, some instructors teach to add the same number on both sides, like this \(2x=7 \to 2x+a=7+a\).

The key is to choose one way and stick with it. Don't try to go back and forth or do both. That's where the confusion comes in.

guideline 2 - - We may multiply by only one, i.e. if we multiply by a number somewhere, we must also divide by the same number.
Just like the first guideline, this can get confusing too, for the exact same reason. The first and best way is to multiply by the same number on one side of the equal sign. It looks like this \(2x=7 \to 2x(b/b)=7\). Notice, that we multiplied and divided by the same number b on the left side.

Of course, we could also multiply both sides of the equal sign by the same number, like this \(2x=7 \to 2x(b)=7(b)\), which is equivalent. The point is to choose one or the other and stick with it. Then, when someone shows you the other way, go back to the way you understand and notice the equivalency.

key to guidelines 1 and 2 - - Now, here is the key to getting your head around the previous two guidelines. If you have an equal sign, think of it as a scale with two sides and the equal sign is the pole in between, like this picture. When you are given an equation in a problem statement, the equation is balanced. Your job is to keep it balanced. The only way to do that is to either not change the weight on one side or by changing the weight on both sides by the same amount.
Of course, if you have no equal sign, your only choice is add/multiply the same number to the equation. So the first technique in both of the first two guidelines works with or without an equal sign. This is another great reason to learn it that way first.

guideline 3 - - Completing the square works only on polynomials whose highest power is \(2\). However, you can sometimes use substitution to get your expression in this form or you can isolate the squared the part of the equation to then be able to use this technique. A quick example of the last idea is if you have an expression like \( 6x^3-36x^2 \), you can factor out an x first to get \( x(6x^2-36x) \) and complete the square on the inside expression.

There are additional guidelines which we explain by working through the next example.

Example - - Complete the square on the expression \( 2x^2-12x \).

First, there must always, always, always, without exception, be a positive one coefficient on the \( x^2 \) term (guideline 4). That means we need to factor out a \(2\) to get \( 2(x^2-6x) \).
Now we take the coefficient of the x-term, which is \(-6\) (notice I took the sign too; you'll see why in a minute) and we divide the \( -6 \) by \(2\) and square it, like this \(\displaystyle{ \left(\frac{-6}{2}\right)^2 = (-3)^2 = 9}\). Okay, then we take that \(9\) and add it and subtract it INSIDE the parentheses to get \( 2(x^2-6x+9-9) \).

Now you are saying to yourself, this instructor is crazy. You didn't really do anything. Aha! That's correct (not about the crazy part but about the part that we didn't do anything). We didn't change the problem. That is important. We don't want to change the problem. By adding and subtracting the same number, we are essentially adding zero. That's good.

Notice we emphasized that both operations had to happen inside the parentheses (guideline 5). Because of the \(2\) that we factored out, if we did one inside and one outside, we would be adding on the inside by \(18\) and subtracting on the outside by \(9\). This would change the problem and our answer would be end up being incorrect. So, we added and subtracted by \(9\), both inside the parentheses.

Okay, now we are going to take the \(9\) we added and associate it with the terms with x in them. Look closely. \( x^2-6x+9=(x-3)^2 \) Because of the way we got the number \(9\) (by dividing by two and squaring) this creates a squared term, thus 'completing' the square. Do you see that?

Now let's see what we have. \( 2(x^2-6x+9-9)=2[(x-3)^2-9] \) We can distribute the \(2\) on the right and end up with \( 2x^2-12x = 2(x-3)^2-18 \). And that's it. (If you need to, take a minute to multiply out the expression on the right side of the equal sign to convince yourself that it is the same as what we started out with.)

Final Answer

\( 2x^2-12x = 2(x-3)^2-18 \)

One question we often get is, how do you know if the sign inside squared term is positive or negative? In the above example, you might ask whether the squared term is \((x-3)\) or \((x+3)\). Well, the sign in the squared term should be the same sign as the coefficient of the x-term in the original problem (after we factor out what we need to get a positive one in front of the squared term). In this case, the coefficient of the x-term is negative (\(-6\)) and so the squared term is \((x-3)\). If you multiply out \(2(x-3)^2-18\) to get \(2x^2-12x\) in the final answer above, you may be able to see this more clearly.

Here is an interesting video that may help you visualize what is going on with this technique.

As a side note, you can use completing the square to derive the quadratic formula. ( This is important to know since knowing where an equation comes from means you really understand when and how to use an equation. ) This next video shows how to do this. Before watching this video, try it on your own.

completing the square → quadratic formula

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Practice Problems

Instructions - - Unless otherwise instructed, complete the square for the following equations. If you are asked to solve an equation (calculate values that satisfy the equation), use completing the square. Give all answers in exact form.

Level A - Basic

Practice A01

\(y=3x^2-2x+1\)

solution

Practice A02

Solve \(x^2-6x+8=0\)

solution

Practice A03

Solve \(x^2+4x-20=12\)

solution

Practice A04

Solve \(a^2+2a-24=0\)

solution

Practice A05

Solve \(x^2+4x-7=0\)

solution

Practice A06

Solve \(3x^2-1=12x\)

solution

Practice A07

Solve \(x^2+5x+6=0\)

solution

Practice A08

Solve \(2x^2+8x-10=0\)

solution

Practice A09

Solve \(x^2+8x+15=0\)

solution

Practice A10

Solve \(-x^2+8x-15=0\)

solution

Practice A11

\(y=x^2+16x-57\)

solution

Practice A12

\(y=-3x^2+24x-27\)

solution


Level B - Intermediate

Practice B01

Solve \(4x^2-10x+1=0\)

solution

Practice B02

Solve \(-5x^2+3x=-2\)

solution

Practice B03

Solve \(2x^2-6x+3=0\)

solution

Practice B04

Solve \(2x^2-7x-15=0\)

solution

Practice B05

Solve \(6x^2-7x-3=0\)

solution

Practice B06

Solve \(4x^2+40x+280=0\)

solution

Practice B07

Solve \(10x^2-30x-8=0\)

solution


Level C - Advanced

Practice C01

Complete the square on both x and y for \(0=x^2+y^2-12x-22y+139\).

answer

solution

Practice C02

Find the value of c that makes \(x^2-44x+c\) a perfect trinomial and write in factored form.

solution

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