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• Practice, practice, practice Math can be an intimidating subject. Each new topic we learn has symbols and problems we have never seen. The unknowing... Read More

• 10.1 Solve Quadratic Equations Using the Square Root Property
• Introduction
• 1.1 Introduction to Whole Numbers
• 1.2 Use the Language of Algebra
• 1.3 Add and Subtract Integers
• 1.4 Multiply and Divide Integers
• 1.5 Visualize Fractions
• 1.6 Add and Subtract Fractions
• 1.7 Decimals
• 1.8 The Real Numbers
• 1.9 Properties of Real Numbers
• 1.10 Systems of Measurement
• Key Concepts
• Review Exercises
• Practice Test
• 2.1 Solve Equations Using the Subtraction and Addition Properties of Equality
• 2.2 Solve Equations using the Division and Multiplication Properties of Equality
• 2.3 Solve Equations with Variables and Constants on Both Sides
• 2.4 Use a General Strategy to Solve Linear Equations
• 2.5 Solve Equations with Fractions or Decimals
• 2.6 Solve a Formula for a Specific Variable
• 2.7 Solve Linear Inequalities
• 3.1 Use a Problem-Solving Strategy
• 3.2 Solve Percent Applications
• 3.3 Solve Mixture Applications
• 3.4 Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem
• 3.5 Solve Uniform Motion Applications
• 3.6 Solve Applications with Linear Inequalities
• 4.1 Use the Rectangular Coordinate System
• 4.2 Graph Linear Equations in Two Variables
• 4.3 Graph with Intercepts
• 4.4 Understand Slope of a Line
• 4.5 Use the Slope-Intercept Form of an Equation of a Line
• 4.6 Find the Equation of a Line
• 4.7 Graphs of Linear Inequalities
• 5.1 Solve Systems of Equations by Graphing
• 5.2 Solving Systems of Equations by Substitution
• 5.3 Solve Systems of Equations by Elimination
• 5.4 Solve Applications with Systems of Equations
• 5.5 Solve Mixture Applications with Systems of Equations
• 5.6 Graphing Systems of Linear Inequalities
• 6.1 Add and Subtract Polynomials
• 6.2 Use Multiplication Properties of Exponents
• 6.3 Multiply Polynomials
• 6.4 Special Products
• 6.5 Divide Monomials
• 6.6 Divide Polynomials
• 6.7 Integer Exponents and Scientific Notation
• 7.1 Greatest Common Factor and Factor by Grouping
• 7.2 Factor Trinomials of the Form x2+bx+c
• 7.3 Factor Trinomials of the Form ax2+bx+c
• 7.4 Factor Special Products
• 7.5 General Strategy for Factoring Polynomials
• 7.6 Quadratic Equations
• 8.1 Simplify Rational Expressions
• 8.2 Multiply and Divide Rational Expressions
• 8.3 Add and Subtract Rational Expressions with a Common Denominator
• 8.4 Add and Subtract Rational Expressions with Unlike Denominators
• 8.5 Simplify Complex Rational Expressions
• 8.6 Solve Rational Equations
• 8.7 Solve Proportion and Similar Figure Applications
• 8.8 Solve Uniform Motion and Work Applications
• 8.9 Use Direct and Inverse Variation
• 9.1 Simplify and Use Square Roots
• 9.2 Simplify Square Roots
• 9.3 Add and Subtract Square Roots
• 9.4 Multiply Square Roots
• 9.5 Divide Square Roots
• 9.6 Solve Equations with Square Roots
• 9.7 Higher Roots
• 9.8 Rational Exponents
• 10.2 Solve Quadratic Equations by Completing the Square
• 10.3 Solve Quadratic Equations Using the Quadratic Formula
• 10.4 Solve Applications Modeled by Quadratic Equations
• 10.5 Graphing Quadratic Equations in Two Variables

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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Solved quadratic equation examples

x^2 -6x + 3 = 0

x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a }

x = \dfrac{ -(-6) \pm \sqrt{(-6)^2 - 4(1)(3)}}{ 2(1) }

x = \dfrac{ 6 \pm \sqrt{36 - 12}}{ 2 }

x = \dfrac{ 6 \pm \sqrt{24}}{ 2 }

  x = \frac{ 6 \pm 2 \sqrt{6}}{ 2 }

x = = 3+\sqrt{6} Or x = 3- \sqrt{6}

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Acceptable Math symbols and their usage If you choose to write your mathematical statements, here is a list of acceptable math symbols and operators.

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Is it Quadratic?

Only if it can be put in the form ax 2 + bx + c = 0 , and a is not zero .

The name comes from "quad" meaning square, as the variable is squared (in other words x 2 ).

These are all quadratic equations in disguise:

How Does this Work?

The solution(s) to a quadratic equation can be calculated using the Quadratic Formula :

The "±" means we need to do a plus AND a minus, so there are normally TWO solutions !

The blue part ( b 2 - 4ac ) is called the "discriminant", because it can "discriminate" between the possible types of answer:

• when it is positive, we get two real solutions,
• when it is zero we get just ONE solution,
• when it is negative we get complex solutions.

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In a short moment, the calculator will show the roots. A quadratic equation can have one of three types of roots: two distinct real roots, one repeated real root, or two complex roots.

What Is a Quadratic Equation?

A quadratic equation is a particular type of polynomial equation of the second degree. It has the following form:

• $$$ x $$$ is the variable or the unknown we aim to find.
• $$$ a $$$ , $$$ b $$$ and $$$ c $$$ are some constants.
• It's essential that $$$ a\ne0 $$$ , otherwise, the equation will not be quadratic but linear.

Features of Quadratic Equations

The graphical representation of a quadratic function is a parabola. Depending on the values of $$$ a $$$ , $$$ b $$$ , and $$$ c $$$ , the parabola can open upwards or downwards and have its vertex anywhere in the coordinate plane.

Using the Quadratic Formula

The quadratic formula is a universal method to find the roots or solutions of any quadratic equation. It is expressed as:

The term inside the square root is called the discriminant. It determines the nature of the roots:

• If $$$ b^2-4ac\gt0 $$$ , the equation has two distinct real roots.
• If $$$ b^2-4ac=0 $$$ , there's one real root (a repeated root).
• If $$$ b^2-4ac\lt0 $$$ , the equation has two complex roots.

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Course: Algebra 1   >   Unit 14

Solving quadratics by taking square roots.

• Quadratics by taking square roots (intro)
• Solving quadratics by taking square roots examples
• Quadratics by taking square roots
• Quadratics by taking square roots: strategy
• Solving quadratics by taking square roots: with steps
• Quadratics by taking square roots: with steps
• Solving simple quadratics review

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