Square Roots and Completing the Square

 

Learning Objective(s)

·         Solve quadratic equations by using the Square Root Property.

·         Identify and complete perfect square trinomials.

·         Solve quadratic equations by completing the square.

 

Introduction

Quadratic equations can be solved in many ways. You may already be familiar with factoring to solve some quadratic equations. However, not all quadratic equations can be factored. In this topic, you will use square roots to learn another way to solve quadratic equations—and this method will work with all quadratic equations.

 

Solving Quadratics Using Square Roots

One way to solve the quadratic equation x2  = 9 is to subtract 9 from both sides to get one side equal to 0: x2  – 9 = 0. The expression on the left can be factored:

(x + 3)(x – 3) = 0. Using the zero factor property, you know this means x + 3 = 0 or x – 3 = 0, so x = 3 or 3.

 

Another property would let you solve that equation more easily.

 

The Square Root Property

 

If x2 = a, then x =  or .

 

The property above says that you can take the square root of both sides of an equation, but you have to think about two cases: the positive square root of a and the negative square root of a.

 

 

A shortcut way to write “” or “” is . The symbol ± is often read “positive or negative.” If it is used as an operation (addition or subtraction), it is read “plus or minus.”

 

 

Example

Problem

Solve using the Square Root Property. x2 = 9

 

x2 = 9

x =

x = ±3

Since one side is simply x2, you can take the square root of both sides to get x on one side. Don’t forget to use both positive and negative square roots!

Answer

x = ±3 (that is, x = 3 or −3)

 

 

Notice that there is a difference here in solving x2 = 9 and finding . For x2 = 9, you are looking for all numbers whose square is 9. For , you only want the principal (nonnegative) square root. The negative of the principal square root is ; both would be . Unless there is a symbol in front of the radical sign, only the nonnegative value is wanted!

 

In the example above, you can take the square root of both sides easily because there is only one term on each side. In some equations, you may need to do some work to get the equation in this form. You will find that this involves isolating x2.

 

 

Example

Problem

Solve. 10x2 + 5 = 85

 

 

10x2 + 5 = 85

 

 

 

 

 

 

If you try taking the square root of both sides of the original equation, you will have  on the left, and you can’t simplify that. Subtract 5 from both sides to get the x2 term by itself.

 

10x2 = 80

 

You could now take the square root of both sides, but you would have

as a coefficient, and you would need to divide by that coefficient. Dividing by 10 before you take the square root will be a little easier.

 

Now you have only x2 on the left, so you can use the Square Root Property easily.

 

Be sure to simplify the radical if possible.

 

 

Answer

 

 

 

Sometimes more than just the x is being squared:

 

 

Example

Problem

Solve. (x – 2)2 – 50 = 0

 

 

(x – 2)2 – 50 = 0

Again, taking the square root of both sides at this stage will leave something you can’t work with on the left. Start by adding 50 to both sides.

 

(x – 2)2 = 50

x – 2 =

Because (x – 2)2 is a squared quantity, you can take the square root of both sides.

 

To isolate x on the left, you need to add 2 to both sides.

 

Be sure to simplify the radical if possible.

 

 

Answer

 

 

 

This method can be helpful when solving real-world problems.

 

 

Example

Problem

The formula for compounding interest annually is

A = P(1 + r)t, where A is the balance after t years, when P is the principal (initial amount invested) and r is the interest rate.

 

Find the interest rate r if $3,000 is invested and grows to $3,307.50 after 2 years.

 

A= P(1 + r)t

A = 3,307.50

t = 2

P = 3,000

First identify what you know. The amount after 2 years is 3,307.50, so

A = 3,307.50. This also means t = 2. The principal P is the original amount invested, so that is 3,000.

 

3,307.50 = 3,000(1 + r)2

Substitute the values for the variables you know. Only r is left, so try to isolate r.

 

Dividing both sides by 3000 leaves only (1 + r)2 on the right. Because (1 + r)2 is a squared quantity you can use the Square Root Property.

 

Don’t forget the ±!

 

±1.05 = 1 + r

Using a calculator, you can find that  is 1.05.

 

±1.05 – 1 = r

Subtract 1 from both sides to isolate r on the right.

 

1.05 – 1 = r, or

−1.05 – 1 = r

You now have two equations, one using 1.05 and one using −1.05.

 

r = 0.05 or −2.05

Simplifying the two equations gives two solutions to the equation.

Answer

The interest rate is 0.05, or 5%.

Notice that a negative interest rate doesn’t make sense for this context, so only the positive value could be the interest rate. The -2.05 is an extraneous solution and must be discarded.

 

 

Solve. (x – 3)2 – 2 = 16

 

A) x = 3 ±

 

B) x = 3 +

 

C) x = 7

 

D) x = 1 or 9

 

Show/Hide Answer

A) x = 3 ±

Correct. Before taking the square root, add 2 to both sides: (x – 3)2 = 18. Applying the Square Root Property gives x – 3 = , so x = 3 ± . Simplifying the radical gives .

 

B) x = 3 +

Incorrect. You forgot the negative square root when you took the square root of both sides. Before taking the square root, add 2 to both sides: (x – 3)2 = 18. Applying the Square Root Property gives x – 3 = , so x = 3 ± . Simplifying the radical gives 3 ± .

 

C) x = 7

Incorrect. There are two mistakes here: Knowing the square root of 16 may have made you forget that to solve this equation, the squared quantity needs to be isolated Before taking the square root, add 2 to both sides: (x – 3)2 = 18. Applying the Square Root Property gives

x – 3 = . (Note that both the positive and negative square roots are included; this is the other probable mistake.) So, x = 3 ± . Simplifying the radical gives 3 ± .

 

D) x = 1 or 9

Incorrect. Knowing the square root of 16 may have made you forget that to solve this equation, the squared quantity needs to be isolated Before taking the square root, add 2 to both sides: (x – 3)2 = 18. Applying the Square Root Property gives x – 3 = . So, x = 3 ± . Simplifying the radical gives 3 ± .

 

 

 

Perfect Square Trinomials

Of course, quadratic equations often will not come in the format of the examples above. Most of them will have x terms. However, you may be able to factor the expression into a squared binomial—and if not, you can still use squared binomials to help you.

 

First, let’s look at squared binomials. Some of the above examples have squared binomials: (1 + r)2 and (x – 2)2 are squared binomials. (They are binomials, two terms, that are squared.) If you expand these, you get a perfect square trinomial. For example, (1 + r)2 = (1 + r)(1 + r) = 1 + 2r + r2, or r2 + 2r + 1. The trinomial r2 + 2r + 1 is a perfect square trinomial. Notice that the first and last terms are squares (r2 and 1). The middle term is twice the product of the square roots of the first and last terms, the square roots are r and 1, and the middle term is 2(r)(1).

 

Perfect square trinomials have the form r2 + 2rs + s2 and can be factored as (r + s)2, or they have the form r2 – 2rs + s2 and can be factored as (rs)2. Let’s factor a perfect square trinomial into a squared binomial.

 

 

Example

Problem

Factor 9x2 – 24x + 16.

 

 

9x2 = (3x)2

16 = 42

First notice that the x2 term and the constant term are both perfect squares.

 

24x = 2(3x)(4)

 

Then notice that the middle term (ignoring the sign) is twice the product of the square roots of the other terms.

 

r = 3x

s = 4

 

9x2 – 24x + 16 = (3x – 4)2

A trinomial in the form r2 – 2rs + s2 can be factored as (rs)2.

 

In this case, the middle term is subtracted, so subtract r and s and square it to get (rs)2.

Answer

(3x – 4)2

 

 

 

You can use the procedure in this next example to help you solve equations where you identify perfect square trinomials, even if the equation is not set equal to 0.

 

 

Example

Problem

Solve. 4x2 + 20x + 25 = 8

 

 

4x2 + 20x + 25 = 8

Since there’s an x term, you can’t use the Square Root Property immediately (or even after adding or dividing by a constant).

 

Notice, however, that the x2 and constant terms on the left are both perfect squares: (2x)2 and 52. Check the middle term: is it 2(2x)(5)? Yes!

 

(2x + 5)2 = 8

 

A trinomial in the form r2 + 2rs + s2 can be factored as (r + s)2, so rewrite the left side as a squared binomial.

 

Now you can use the Square Root Property. Some additional steps are needed to isolate x.

 

Simplify the radical when possible.

 

Answer

 

 

 

 

You may have looked at the problem above and thought “why not subtract 8 from both sides first, making the equation 4x2 + 20x + 17 = 0?” This is one technique for solving for x—the problem is that even if you did this, the equation 4x2 + 20x + 17 = 0 cannot be factored with real numbers. (Try it—can you think of two numbers whose product is 68 and whose sum is 20?)

 

Since the expression 4x2 + 20x + 25 can be identified as a perfect square trinomial, it is best to factor it as (2x + 5)2 and then use the Square Root Property.

 

Completing the Square

One way to solve quadratic equations is by completing the square. When you don’t have a perfect square trinomial, you can create one by adding a constant term that is a perfect square to both sides of the equation. Let’s see how to find that constant term.

 

“Completing the square” does exactly what it says—it takes something that is not a square and makes it one. This idea can be illustrated using an area model of the binomial x2 + bx.

 

 

In this example, the area of the overall rectangle is given by x(x + b).

 

Now let's make this rectangle into a square. First, divide the red rectangle with area bx into two equal rectangles each with area . Then rotate and reposition one of them. You haven't changed the size of the red area—it still adds up to bx.

 

 

 

 

The red rectangles now make up two sides of a square, shown in white. The area of that square is the length of the red rectangles squared, or .

 

Here comes the cool part—do you see that when the white square is added to the blue and red regions, the whole shape is also now a square? In other words, you've "completed the square!" By adding the quantity  to the original binomial, you've made a square, a square with sides of .

 

Notice that the area of this square can be written as a squared binomial: .

 

 

Finding a Value that will Complete the Square in an Expression

 

To complete the square for an expression of the form x2 + bx:

·         Identify the value of b;

·         Calculate and add .

The expression becomes .

 

 

 

Example

Problem

Find the number to add to x2 + 8x to make it a perfect square trinomial.

 

x2 + 8x

b = 8

 

First identify b if this has the form x2 + bx.

 

 

 

 

To complete the square, add .

 

b = 8, so

 

x2 + 8x + (4)2

x2 + 8x + 16

 

x2 + 8x + 16 = (x + 4)2

 

Simplify.

 

Check that the result is a perfect square trinomial. (x + 4)2 =

x2 + 4x + 4x + 16 =

x2 + 8x + 16, so it is.

Answer

Adding +16 will make x2 + 8x a perfect square trinomial.

 

 

Notice that  is always positive, since it is the square of a number. When you complete the square, you are always adding a positive value.

 

 

Use completing the square to find the value to add that makes x2 – 12x a perfect square trinomial. Then write the expression as the square of a binomial.

 

A) add 12; (x – 6)2

B) add 36; (x + 6)2

C) add −12; (x – 12)2

D) add 36; (x – 6)2

 

Show/Hide Answer

A) add 12; (x – 6)2

Incorrect. The value of b is −12, so the value to add is , not 12. Add 36 to get x2 – 12x + 36. The correct answer is (x – 6)2.

 

B) add 36; (x + 6)2

Incorrect. The value to add has been calculated correctly: . However, the resulting trinomial is x2 – 12x + 36, which factors to (x – 6)2, not (x + 6)2. The correct answer is (x – 6)2.

 

C) add −12; (x – 12)2

Incorrect. The value of b is −12, so the value to add is , not −12. Note also, that the number you add will always be positive because it is the square of a number. The correct answer is (x – 6)2.

 

D) add 36; (x – 6)2

Correct. The value to add is  and the resulting trinomial x2 – 12x + 36 factors to (x – 6)2.

 

 

 

Solving a Quadratic Equation using Completing the Square

 

You can use completing the square to help you solve a quadratic equation that cannot be solved by factoring.

 

Let’s start by seeing what happens when you complete the square in an equation. In the example below, notice that completing the square will result in adding a number to both sides of the equation—you have to do this in order to keep both sides equal!

 

 

Example

Problem

Rewrite x2 + 6x = 8 so that the left side is a perfect square trinomial.

 

x2 + 6x = 8

b = 6

 

This equation has a constant of 8. Ignore it for now and focus on the x2 and x terms on the left side of the equation. The left side has the form x2 + bx, so you can identify b.

 

 

x2 + 6x + 9 = 8 + 9

 

To complete the square, add  to the left side.

 

b = 6, so

 

This is an equation, though, so you must add the same number to the right side as well.

 

 

x2 + 6x + 9 = 17

 

x2 + 6x + 9 = 17

(x + 3)2 = 17

 

Simplify.

 

Check that the left side is a perfect square trinomial.

(x + 3)2 =

x2 + 3x + 3x + 9 =

x2 + 6x + 9, so it is.

Answer

x2 + 6x + 9 = 17

 

 

 

Can you see that completing the square in an equation is very similar to completing the square in an expression? The main difference is that you have to add the new number (+9 in this case) to both sides of the equation to maintain equality.

 

Now let’s look at an example where you are using completing the square to actually solve an equation, finding a value for the variable.

 

 

Example

Problem

Solve. x2 – 12x – 4 = 0

 

 

x2 – 12x = 4

b = −12

 

 

 

Since you cannot factor the trinomial on the left side, you will use completing the square to solve the equation.

 

Rewrite the equation with the left side in the form
x2 + bx, to prepare to complete the square. Identify b.

 

 

 

x2 – 12x + 36 = 4 + 36

 

x2 – 12x + 36 = 40

 

Figure out what value to add to complete the square. Add  to complete the square, so = .

 

Add the value to both sides of the equation and simplify.

 

(x – 6)2 =  40

 

 

Rewrite the left side as a squared binomial.

 

 

 

 

Use the Square Root Property. Remember to include both the positive and negative square root, or you’ll miss one of the solutions.

 

 

Solve for x by adding 6 to both sides. Simplify as needed.

 

 

Answer

 

 

 

You may have noticed that because you have to use both square roots, all the examples have two solutions. Here is another example that’s slightly different.

 

 

Example

Problem

Solve. x2 + 16x + 17 = −47

 

x2 + 16x = −64

b = 16

 

Rewrite the equation so the left side has the form x2 + bx. Identify b.

 

 

x2 + 16x + 64 = −64 + 64

x2 + 16x + 64 = 0

 

Add , which is , to both sides.

 

 

(x + 8)2 = 0

 

Write the left side as a squared binomial.

 

x + 8 = 0

 

 

 

Take the square roots of both sides. Normally both positive and negative square roots are needed, but 0 is neither positive nor negative. 0 has only one root.

 

x = −8

 

Solve for x.

Answer

x = −8

 

 

 

Take a closer look at this problem and you may see something familiar. Instead of completing the square, try adding 47 to both sides in the equation. The equation x2 + 16x + 17 = −47 becomes x2 + 16x + 64 = 0. Can you factor this equation using grouping? (Think of two numbers whose product is 64 and whose sum is 16).

 

It can be factored as (x + 8)(x + 8) = 0, of course! Knowing how to complete the square is very helpful, but it is not always the only way to solve an equation.

 

 

Solve. x2 – 16x = −1

 

A)

B)

C)

D)

 

Show/Hide Answer

A)

Incorrect. You appear to have added 16 and then factored the trinomial incorrectly. However, the value of b is −16, so to complete the square, add  (not 16). Add 64 to both sides to get x2 – 16x + 64 = 63. This is (x – 8)2 = 63, so  That means  The correct answer is .

 

B)

Incorrect. The value to add has been calculated correctly:  However, the resulting trinomial is x2 – 16x + 64, which factors to (x – 8)2, not (x + 8)2. The equation becomes (x – 8)2 = 63, so  That means  The correct answer is .

 

C)

Incorrect. The value to add has been calculated correctly:  However, you appear to have added 64 to the left side of the equation only, and not to the right side. The equation becomes (x – 8)2 = 63, so  That means  The correct answer is .

 

D)

Correct. The value of b is −16, so to complete the square, add . Add 64 to both sides to get x2 – 16x + 64 = 63. This is (x – 8)2 = 63, so  That means

 

 

 

Summary

Completing the square is used to change a binomial of the form x2 + bx into a perfect square trinomial , which can be factored to . When solving quadratic equations by completing the square, be careful to add  to both sides of the equation to maintain equality. The Square Root Property can then be used to solve for x. With the Square Root Property, be careful to include both the principal square root and its opposite. Be sure to simplify as needed.