SAT Math: Completing the Square

When the new SAT came out in March 2016, it became clear that the test makers were looking to test some skills that hadn’t appeared on the previous version of the test. One of those skills is “completing the square,” which can be helpful when you are working on problems involving circles and parabolas.

So what is completing the square? Well, a perfect square trinomial x² + bx + c is one that can be rewritten in the form (x + k)², where k is some integer. For instance x² + 10x + 25 is a perfect square trinomial because is can be factored into (x + 5)(x + 5) = (x + 5)².

When you are asked to complete the square, you need to find that value of c (there is always only one possible value) that will make the trinomial a perfect square. For instance, suppose you were asked this question:

What number will complete the square for x²  + 6x + ___ ?

In other words, we want to know what number to put in the blank so that we “complete the square” and create a perfect square trinomial. Finding that number is fairly easy. We simply cut the “b” number (the coefficient of x) in half and square the result. In this case, half of 6 is 3 and 3 squared is 9. The  resulting polynomial, x² + 6x + 9, can be factored into (x + 3)(x + 3) or (x + 3)².

Let’s take a look at a few problems where we might need to use this technique.

The equation below represents a circle. What are the center and the radius of the circle?

x² + y² + 10x – 12y + 45 = 0

A)  (-5, 6); r = 4
B)  (-5, 6); r = 16
C)  (5, -6); r = 4
D)  (5, -6); r = 16

Recall that the equation of a circle is (x – h)² + (y – k)² = r². So we need to make the given equation look like this. We can do this by completing the square twice, once for each variable. Begin by rearranging the terms and leaving blanks for the numbers we’ll need to add to complete the square. At the same time, move the constant term 45 to the other side of the equation by subtracting 45 on each side.

x² + 10x + ___ + y² – 12y + ___ = -45

Now complete the square to find the number that goes in each blank. Remember that if you add a number to one side of the equation, you need to add it to the other as well to keep the equation “in balance.”

x² + 10x + ___ + y² – 12y + ___ = -45
x² + 10x + 25 + y² – 12y + 36 = -45 + 25 + 36
x² + 10x + 25 + y² – 12y + 36 = 16

Now factor each of the two trinomials you’ve created to convert the equation to the form we’re looking for.

(x² + 10x + 25) + (y² – 12y + 36) = 16
(x + 5)(x + 5) + (y – 6)(y – 6) = 16
(x + 5)² + (y – 6)² = 4²

So, the center of the circle is (-5, 6) and the radius of the circle is 4. The correct answer choice is A.

Here’s a second problem, this one involving parabolas.

What is the vertex of the graph of f(x) = x²8x + 9?

A)  (-8, 9)
B)  (4, -7)
C)  (4, 9)
D)  (8, -9)

Recall that the equation of a parabola can be written in vertex form: f(x) = (x – h)² + k. In this form, the vertex of the graph is located at the point (h, k). Let’s see if we can make the given equation look like this by completing the square. In this instance, instead of moving the constant to the other side of the equation, we’ll keep it on the same side. We’ll also “balance” the equation by subtracting the number we add in the blank from the same side of the equation. 

 f(x) = x²8x + 9
f(x) = x²8x + ___ + 9
f(x) = x²8x + 16 + 9 – 16
f(x) = (x²8x + 16) – 7
f(x) = (x – 4)(x – 4) – 7
f(x) = (x – 4)² – 7

The vertex of the parabola is (4, -7) and the correct answer is Choice B.

(Note: If you know quadratic equations and parabolas well, you may realize that you could have done this problem in a different way. Find the equation of the axis of symmetry using the formula x = -b/2a. This gives you the x-coordinate of the vertex;  then run that value through the equation to find the y-coordinate of the vertex.)

If you’d like to practice some of these problems on your own, go here.

If you have questions about “completing the square” or anything else to do with the new SAT, drop us an email at info@cardinalec.com .

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