unit 4 homework 3 solving quadratic by square roots

Solving Quadratics by Square Root Method

How to solve quadratic equations using the square root method.

This is the “best” method whenever the quadratic equation  only contains [latex]{x^2}[/latex] terms. That implies no presence of any [latex]x[/latex] term being raised to the first power somewhere in the equation.

The general approach is to collect all [latex]{x^2}[/latex] terms on one side of the equation while keeping the constants to the opposite side. After doing so, the next obvious step is to take the square roots of both sides to solve for the value of [latex]x[/latex]. Always attach the [latex] \pm [/latex] symbol when you get the square root of the constant.

Examples of How to Solve Quadratic Equations by Square Root Method

Example 1 : Solve the quadratic equation below using the Square Root Method.

I will isolate the only [latex]{x^2}[/latex] term on the left side by adding both sides by [latex] + 1[/latex]. Then solve the values of [latex]x[/latex] by taking the square roots of both sides of the equation. As I mentioned before, we need to attach the plus or minus symbol to the square root of the constant.

So I have [latex]x = 5[/latex] and [latex]x = – \,5[/latex] as final answers since both of these values satisfy the original quadratic equation. I will leave it to you to verify.

Example 2 : Solve the quadratic equation below using the Square Root Method.

This problem is very similar to the previous example. The only difference is that after I have separated the [latex]{x^2}[/latex] term and the constant in the opposite sides of the equation, I need to divide the equation by the coefficient of the squared term before taking the square roots of both sides.

The final answers are [latex]x = 4[/latex] and [latex]x = – \,4[/latex].

Example 3 : Solve the quadratic equation below using the Square Root Method.

I can see that I have two [latex]{x^2}[/latex] terms, one on each side of the equation. My approach is to collect all the squared terms of [latex]x[/latex] to the left side, and combine all the constants to the right side. Then solve for [latex]x[/latex] as usual, just like in Examples 1 and 2.

The solutions to this quadratic formula are [latex]x = 3[/latex] and [latex]x = – \,3[/latex].

Example 4 : Solve the quadratic equation below using the Square Root Method.

The two parentheses should not bother you at all. The fact remains that all variables come in the squared form, which is what we want. This problem is perfectly solvable using the square root method.

So my first step is to eliminate both of the parentheses by applying the distributive property of multiplication. Once they are gone, I can easily combine like terms. Keep the [latex]{x^2}[/latex] terms to the left, and constants to the right. Finally, apply square root operation in both sides and we’re done!

Not too bad, right?

Example 5 : Solve the quadratic equation below using the Square Root Method.

Since the [latex]x[/latex]-term is being raised to the second power twice, that means, I need to perform two square root operations in order to solve for [latex]x[/latex].

The first step is to have something like this: (   ) 2 = constant . This allows me to get rid of the exponent of the parenthesis on the first application of square root operation.

After doing so, what remains is the “stuff” inside the parenthesis which has an [latex]{x^2}[/latex] term. Well, this is great since I already know how to handle it just like the previous examples.

There’s an [latex]x[/latex]-squared term left after the first application of square root.

Now we have to break up [latex]{x^2} = \pm \,6 + 10[/latex] into two cases because of the “plus” or “minus” in [latex]6[/latex].

• Solve the first case where [latex]6[/latex] is positive .

• Solve the second case where [latex]6[/latex] is negative .

The solutions to this quadratic equations are [latex]x = 4[/latex], [latex]x = – \,4[/latex], [latex]x = 2[/latex], and [latex]x = – \,2[/latex]. Yes, we have four values of [latex]x[/latex] that can satisfy the original quadratic equation.

Example 6 : Solve the quadratic equation below using the Square Root Method.

Example 7 : Solve the quadratic equation below using the Square Root Method.

You might also like these tutorials:

• Solving Quadratic Equations by Factoring Method
• Solving Quadratic Equations by the Quadratic Formula
• Solving Quadratic Equations by Completing the Square

10.1 Solve Quadratic Equations Using the Square Root Property

Learning objectives.

By the end of this section, you will be able to:

• Solve quadratic equations of the form a x 2 = k a x 2 = k using the Square Root Property
• Solve quadratic equations of the form a ( x − h ) 2 = k a ( x − h ) 2 = k using the Square Root Property

Be Prepared 10.1

Before you get started, take this readiness quiz.

Simplify: 75 75 . If you missed this problem, review Example 9.12 .

Be Prepared 10.2

Simplify: 64 3 64 3 . If you missed this problem, review Example 9.67 .

Be Prepared 10.3

Factor: 4 x 2 − 12 x + 9 4 x 2 − 12 x + 9 . If you missed this problem, review Example 7.43 .

Quadratic equations are equations of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 , where a ≠ 0 a ≠ 0 . They differ from linear equations by including a term with the variable raised to the second power. We use different methods to solve quadratic equation s than linear equations, because just adding, subtracting, multiplying, and dividing terms will not isolate the variable.

We have seen that some quadratic equations can be solved by factoring. In this chapter, we will use three other methods to solve quadratic equations.

Solve Quadratic Equations of the Form ax 2 = k Using the Square Root Property

We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation x 2 = 9 x 2 = 9 .

We can easily use factoring to find the solutions of similar equations, like x 2 = 16 x 2 = 16 and x 2 = 25 x 2 = 25 , because 16 and 25 are perfect squares. But what happens when we have an equation like x 2 = 7 x 2 = 7 ? Since 7 is not a perfect square, we cannot solve the equation by factoring.

These equations are all of the form x 2 = k x 2 = k . We defined the square root of a number in this way:

This leads to the Square Root Property .

Square Root Property

If x 2 = k x 2 = k , and k ≥ 0 k ≥ 0 , then x = k or x = − k x = k or x = − k .

Notice that the Square Root Property gives two solutions to an equation of the form x 2 = k x 2 = k : the principal square root of k k and its opposite. We could also write the solution as x = ± k x = ± k .

Now, we will solve the equation x 2 = 9 x 2 = 9 again, this time using the Square Root Property.

What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation x 2 = 7 x 2 = 7 .

Example 10.1

Solve: x 2 = 169 x 2 = 169 .

Use the Square Root Property. Simplify the radical. x 2 = 169 x = ± 169 x = ± 13 Rewrite to show two solutions. x = 13 , x = −13 Use the Square Root Property. Simplify the radical. x 2 = 169 x = ± 169 x = ± 13 Rewrite to show two solutions. x = 13 , x = −13

Try It 10.1

Solve: x 2 = 81 x 2 = 81 .

Try It 10.2

Solve: y 2 = 121 y 2 = 121 .

Example 10.2

How to solve a quadratic equation of the form a x 2 = k a x 2 = k using the square root property.

Solve: x 2 − 48 = 0 x 2 − 48 = 0 .

Try It 10.3

Solve: x 2 − 50 = 0 x 2 − 50 = 0 .

Try It 10.4

Solve: y 2 − 27 = 0 y 2 − 27 = 0 .

Solve a quadratic equation using the Square Root Property.

• Step 1. Isolate the quadratic term and make its coefficient one.
• Step 2. Use Square Root Property.
• Step 3. Simplify the radical.
• Step 4. Check the solutions.

To use the Square Root Property, the coefficient of the variable term must equal 1. In the next example, we must divide both sides of the equation by 5 before using the Square Root Property.

Example 10.3

Solve: 5 m 2 = 80 5 m 2 = 80 .

Try It 10.5

Solve: 2 x 2 = 98 2 x 2 = 98 .

Try It 10.6

Solve: 3 z 2 = 108 3 z 2 = 108 .

The Square Root Property started by stating, ‘If x 2 = k x 2 = k , and k ≥ 0 k ≥ 0 ’. What will happen if k < 0 k < 0 ? This will be the case in the next example.

Example 10.4

Solve: q 2 + 24 = 0 q 2 + 24 = 0 .

Try It 10.7

Solve: c 2 + 12 = 0 c 2 + 12 = 0 .

Try It 10.8

Solve: d 2 + 81 = 0 d 2 + 81 = 0 .

Remember, we first isolate the quadratic term and then make the coefficient equal to one.

Example 10.5

Solve: 2 3 u 2 + 5 = 17 2 3 u 2 + 5 = 17 .

Try It 10.9

Solve: 1 2 x 2 + 4 = 24 1 2 x 2 + 4 = 24 .

Try It 10.10

Solve: 3 4 y 2 − 3 = 18 3 4 y 2 − 3 = 18 .

The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator.

Example 10.6

Solve: 2 c 2 − 4 = 45 2 c 2 − 4 = 45 .

Try It 10.11

Solve: 5 r 2 − 2 = 34 5 r 2 − 2 = 34 .

Try It 10.12

Solve: 3 t 2 + 6 = 70 3 t 2 + 6 = 70 .

Solve Quadratic Equations of the Form a ( x − h ) 2 = k Using the Square Root Property

We can use the Square Root Property to solve an equation like ( x − 3 ) 2 = 16 ( x − 3 ) 2 = 16 , too. We will treat the whole binomial, ( x − 3 ) ( x − 3 ) , as the quadratic term.

Example 10.7

Solve: ( x − 3 ) 2 = 16 ( x − 3 ) 2 = 16 .

Try It 10.13

Solve: ( q + 5 ) 2 = 1 ( q + 5 ) 2 = 1 .

Try It 10.14

Solve: ( r − 3 ) 2 = 25 ( r − 3 ) 2 = 25 .

Example 10.8

Solve: ( y − 7 ) 2 = 12 ( y − 7 ) 2 = 12 .

Try It 10.15

Solve: ( a − 3 ) 2 = 18 ( a − 3 ) 2 = 18 .

Try It 10.16

Solve: ( b + 2 ) 2 = 40 ( b + 2 ) 2 = 40 .

Remember, when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.

Example 10.9

Solve: ( x − 1 2 ) 2 = 5 4 . ( x − 1 2 ) 2 = 5 4 .

Try It 10.17

Solve: ( x − 1 3 ) 2 = 5 9 . ( x − 1 3 ) 2 = 5 9 .

Try It 10.18

Solve: ( y − 3 4 ) 2 = 7 16 . ( y − 3 4 ) 2 = 7 16 .

We will start the solution to the next example by isolating the binomial.

Example 10.10

Solve: ( x − 2 ) 2 + 3 = 30 ( x − 2 ) 2 + 3 = 30 .

Try It 10.19

Solve: ( a − 5 ) 2 + 4 = 24 ( a − 5 ) 2 + 4 = 24 .

Try It 10.20

Solve: ( b − 3 ) 2 − 8 = 24 ( b − 3 ) 2 − 8 = 24 .

Example 10.11

Solve: ( 3 v − 7 ) 2 = −12 ( 3 v − 7 ) 2 = −12 .

Try It 10.21

Solve: ( 3 r + 4 ) 2 = −8 ( 3 r + 4 ) 2 = −8 .

Try It 10.22

Solve: ( 2 t − 8 ) 2 = −10 ( 2 t − 8 ) 2 = −10 .

The left sides of the equations in the next two examples do not seem to be of the form a ( x − h ) 2 a ( x − h ) 2 . But they are perfect square trinomials, so we will factor to put them in the form we need.

Example 10.12

Solve: p 2 − 10 p + 25 = 18 p 2 − 10 p + 25 = 18 .

The left side of the equation is a perfect square trinomial. We will factor it first.

Try It 10.23

Solve: x 2 − 6 x + 9 = 12 x 2 − 6 x + 9 = 12 .

Try It 10.24

Solve: y 2 + 12 y + 36 = 32 y 2 + 12 y + 36 = 32 .

Example 10.13

Solve: 4 n 2 + 4 n + 1 = 16 4 n 2 + 4 n + 1 = 16 .

Again, we notice the left side of the equation is a perfect square trinomial. We will factor it first.

Try It 10.25

Solve: 9 m 2 − 12 m + 4 = 25 9 m 2 − 12 m + 4 = 25 .

Try It 10.26

Solve: 16 n 2 + 40 n + 25 = 4 16 n 2 + 40 n + 25 = 4 .

Access these online resources for additional instruction and practice with solving quadratic equations:

• Solving Quadratic Equations: Solving by Taking Square Roots
• Using Square Roots to Solve Quadratic Equations
• Solving Quadratic Equations: The Square Root Method

Section 10.1 Exercises

Practice makes perfect.

Solve Quadratic Equations of the form a x 2 = k a x 2 = k Using the Square Root Property

In the following exercises, solve the following quadratic equations.

a 2 = 49 a 2 = 49

b 2 = 144 b 2 = 144

r 2 − 24 = 0 r 2 − 24 = 0

t 2 − 75 = 0 t 2 − 75 = 0

u 2 − 300 = 0 u 2 − 300 = 0

v 2 − 80 = 0 v 2 − 80 = 0

4 m 2 = 36 4 m 2 = 36

3 n 2 = 48 3 n 2 = 48

x 2 + 20 = 0 x 2 + 20 = 0

y 2 + 64 = 0 y 2 + 64 = 0

2 5 a 2 + 3 = 11 2 5 a 2 + 3 = 11

3 2 b 2 − 7 = 41 3 2 b 2 − 7 = 41

7 p 2 + 10 = 26 7 p 2 + 10 = 26

2 q 2 + 5 = 30 2 q 2 + 5 = 30

Solve Quadratic Equations of the Form a ( x − h ) 2 = k a ( x − h ) 2 = k Using the Square Root Property

( x + 2 ) 2 = 9 ( x + 2 ) 2 = 9

( y − 5 ) 2 = 36 ( y − 5 ) 2 = 36

( u − 6 ) 2 = 64 ( u − 6 ) 2 = 64

( v + 10 ) 2 = 121 ( v + 10 ) 2 = 121

( m − 6 ) 2 = 20 ( m − 6 ) 2 = 20

( n + 5 ) 2 = 32 ( n + 5 ) 2 = 32

( r − 1 2 ) 2 = 3 4 ( r − 1 2 ) 2 = 3 4

( t − 5 6 ) 2 = 11 25 ( t − 5 6 ) 2 = 11 25

( a − 7 ) 2 + 5 = 55 ( a − 7 ) 2 + 5 = 55

( b − 1 ) 2 − 9 = 39 ( b − 1 ) 2 − 9 = 39

( 5 c + 1 ) 2 = −27 ( 5 c + 1 ) 2 = −27

( 8 d − 6 ) 2 = −24 ( 8 d − 6 ) 2 = −24

m 2 − 4 m + 4 = 8 m 2 − 4 m + 4 = 8

n 2 + 8 n + 16 = 27 n 2 + 8 n + 16 = 27

25 x 2 − 30 x + 9 = 36 25 x 2 − 30 x + 9 = 36

9 y 2 + 12 y + 4 = 9 9 y 2 + 12 y + 4 = 9

Mixed Practice

In the following exercises, solve using the Square Root Property.

2 r 2 = 32 2 r 2 = 32

4 t 2 = 16 4 t 2 = 16

( a − 4 ) 2 = 28 ( a − 4 ) 2 = 28

( b + 7 ) 2 = 8 ( b + 7 ) 2 = 8

9 w 2 − 24 w + 16 = 1 9 w 2 − 24 w + 16 = 1

4 z 2 + 4 z + 1 = 49 4 z 2 + 4 z + 1 = 49

a 2 − 18 = 0 a 2 − 18 = 0

b 2 − 108 = 0 b 2 − 108 = 0

( p − 1 3 ) 2 = 7 9 ( p − 1 3 ) 2 = 7 9

( q − 3 5 ) 2 = 3 4 ( q − 3 5 ) 2 = 3 4

m 2 + 12 = 0 m 2 + 12 = 0

n 2 + 48 = 0 n 2 + 48 = 0

u 2 − 14 u + 49 = 72 u 2 − 14 u + 49 = 72

v 2 + 18 v + 81 = 50 v 2 + 18 v + 81 = 50

( m − 4 ) 2 + 3 = 15 ( m − 4 ) 2 + 3 = 15

( n − 7 ) 2 − 8 = 64 ( n − 7 ) 2 − 8 = 64

( x + 5 ) 2 = 4 ( x + 5 ) 2 = 4

( y − 4 ) 2 = 64 ( y − 4 ) 2 = 64

6 c 2 + 4 = 29 6 c 2 + 4 = 29

2 d 2 − 4 = 77 2 d 2 − 4 = 77

( x − 6 ) 2 + 7 = 3 ( x − 6 ) 2 + 7 = 3

( y − 4 ) 2 + 10 = 9 ( y − 4 ) 2 + 10 = 9

Everyday Math

Paola has enough mulch to cover 48 square feet. She wants to use it to make three square vegetable gardens of equal sizes. Solve the equation 3 s 2 = 48 3 s 2 = 48 to find s s , the length of each garden side.

Kathy is drawing up the blueprints for a house she is designing. She wants to have four square windows of equal size in the living room, with a total area of 64 square feet. Solve the equation 4 s 2 = 64 4 s 2 = 64 to find s s , the length of the sides of the windows.

Writing Exercises

Explain why the equation x 2 + 12 = 8 x 2 + 12 = 8 has no solution.

Explain why the equation y 2 + 8 = 12 y 2 + 8 = 12 has two solutions.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ If most of your checks were:

…confidently: Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help: This must be addressed quickly because topics you do not master become potholes in your road to success. In math, every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no-I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

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Unit 4: Solving Quadratic Equations

Why are we studying this.

In this unit of study, students will continue to deepen their knowledge and understand of quadratic functions. Students will learn how to solve quadratic equations algebraically and interpret them in terms of the graph and in context.

Unit Schedule and Assignments

Weekly online, notebook: unit 4, table of contents & pages, hw #8: due 11/30, hw #10: due 12/14, hw #9: due 12/7, hw #11: due 12/21.

Daily Homework Assignments (w/Answer Keys)

Graphing: practice page.

Factoring Quadratic Expressions: Practice Page

8.1: #3-19 ODD

8.2: #1-15 odd, 9.1: #4-15 all + 18, 19, 22, 9.2: #1-15 odd + 23, 24, 9.3: #1-13 odd + 15-20 all, unit assessments, unit quizzes.

Mid-Unit Quiz #1: Solving Quadratics by Graphing and Factoring - Tuesday, December 4th (Per. 1 & 3) and Wednesday, December 5th (Per. 2)

Mid-Unit Quiz #2: Solving Quadratics by Square Roots, Completing the Square, and Quadratic Formula (POP QUIZ)

Unit 4 Assessment: Tuesday, December 18th (Per. 1 & 3) and Wednesday, December 19th (Per. 2)

Solving Quadratic Equations by Graphing

Unit Foldable: Solving Quadratic Equations

In Class Slides

Practice Page

Factoring Quadratic Expressions

In Class Slides (Day 1)

Factoring Trinomials Match Up

Factoring Quadratic Expressions (Orduna - video)

8.1-8.2: Solving Quadratic Equations by Factoring

8.1 Textbook Pages

8.2 Textbook Pages

In Class Slides (Day 2)

Notes/Practice Page

Solving Quadratic Equations by Factoring (Orduna - video)

For extra tutorial videos, visit my.hrw.com and check out...

8.1: Math on the Spot Video

8.2: Math on the Spot Video

9.1: Solving Quadratic Equations by Square Roots

9.1 Textbook Pages

Notes/Practice Pages

Choose Your Own Adventure

9.2: Solving Quadratic Equations by Completing the Square

9.2 Textbook Pages

Coloring Page

9.3: Solving Quadratics by Quadratic Formula

9.3 Textbook Pages

Quadratic Formula Tic-Tac-Toe

UNIT REVIEW RESOURCES

Word Problem Practice

Solving by Different Methods

Quadratics Puzzle

4 to 1 Review Game

Quadratic Equation Worksheets (pdfs)

Free worksheets with answer keys.

Enjoy these free sheets. Each one has model problems worked out step by step, practice problems, as well as challenge questions at the sheets end. Plus each one comes with an answer key.

• Solve Quadratic Equations by Factoring
• Solve Quadratic Equations by Completing the Square
• Quadratic Formula Worksheet (real solutions)
• Quadratic Formula Worksheet (complex solutions)
• Quadratic Formula Worksheet (both real and complex solutions)
• Discriminant Worksheet
• Sum and Product of Roots
• Radical Equations Worksheet

Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there!

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