Algebra1 Solving Radical Equations

CONFIDENTIAL 1

Algebra1Algebra1

Solving RadicalSolving RadicalEquationsEquations

CONFIDENTIAL 2

Warm UpWarm Up

1) A dessert menu offers 6 different selections. The restaurant offers a dessert sampler that includes small portions of any 4 different choices from the

dessert menu. How many different dessert samplers are possible?

CONFIDENTIAL 3

Solving Radical EquationsSolving Radical Equations

A radical equation is an equation thatcontains a variable within a radical. Inthis course, you will only study radicalequations that contain square roots.

Recall that you use inverse operations to solve equations. For nonnegative numbers,

squaring and taking the square root are inverse operations. When an equation

contains a variable within a square root, square both sides of the equation to solve.

CONFIDENTIAL 4

Power Property of Power Property of EqualityEquality

WORDS NUMBERS ALGEBRA

You can square both sides of

an equation, and the resulting

equation is still true.

3 = 1 + 2

(3)2 + (1 + 2)2

9 = 9

If a and b are real numbers

and a = b,

then a2 = b2.

CONFIDENTIAL 5

A) √x = 8

√(x)2 = 82

x = 64

Square both sides.

Check:

Substitute 64 for x in the original equation.

Simplify.

Solving Simple Radical EquationsSolving Simple Radical Equations

Solve each equation. Check your answer.

√x = 8 √(64) 8 8 8

CONFIDENTIAL 6

B) 6 = √(4x)

√62 = √(4x)2

36 = x

9 = x

Square both sides.

Check:


Simplify.

6 = √(4x) 6 √(4(9)) 6 √(36) 6 6

Divide both sides by 4.

CONFIDENTIAL 7

Now you try!

Multiply. Write each product in simplest form.

1a) √x = 6

1b) 9 = √(27x)

1c) √(3x) = 1

CONFIDENTIAL 8

Some square-root equations do not have the square root isolated.

To solve theseequations, you may have to

isolate the square root before squaring both sides.

You can do this by using one or more inverse operations.

CONFIDENTIAL 9

Solving Radical Equations by Adding or SubtractingSolving Radical Equations by Adding or Subtracting


A) √x + 3 = 10

√x = 7

√(x)2 = 72

x = 49

Subtract 3 from both sides.

Square both sides.

Check:


Simplify.

√x + 3 = 10 √(49) + 3 10 7 + 3 10 10 10

CONFIDENTIAL 10

B) √(x – 5) = 4

√(x – 5)2 = 42

(x – 5) =16

x = 21

Add 5 to both sides.

Square both sides.

Check:


Simplify.

√(x – 5) = 4 √(21 – 5) 4 √(16) 4 4 4

CONFIDENTIAL 11

C) √(2x – 1) + 4 = 7

√(2x – 1) = 3(√(2x – 1))2 = (3)2

2x - 1 = 92x = 10 x = 5

Square both sides.Subtract 4 from both sides.

Check:Substitute 5 for x in the original equation.

Simplify.

√(2x – 1) + 4 = 7 √(2(5) – 1) + 4 7 √(10 - 1) + 4 7 √9 + 4 7 3 + 4 7 7 7

Divide both sides by 2.Add 1 to both sides.

CONFIDENTIAL 12

Now you try!


2a) √x - 2 = 1

2b) √(x + 7) = 5

2c) √(3x + 7) - 1 = 3

CONFIDENTIAL 13

Solving Radical Equations by Multiplying or Solving Radical Equations by Multiplying or DividingDividingSolve each equation. Check your answer.

A) 3√x = 21

Method 1:√x = 7√(x)2 = 72

x = 49

Method 2:3√x = 21(3√(x))2 = (21)2

9x = 441 x = 49


Square both sides.


Simplify.

3√x = 21 3√(49) 21 3(7) 21 21 21


Square both sides.

CONFIDENTIAL 14

B) √x = 5 3 Method 1:√x = 15√(x)2 = (15)2

x = 225

Method 2:√x = (5)2

3 x = 25 9 x = 225

Multiply both sides by 3.

Square both sides.


Simplify.

√x = 5 3 √(225) 5 3 15 5 3 5 5

Multiply both sides by 9.

Square both sides.

CONFIDENTIAL 15

Now you try!


3a) 2√x = 22

3b) 2 = √x 4

3c) 2√x = 4 5

CONFIDENTIAL 16

Solving Radical Equations with Solving Radical Equations with Square Roots on Both SidesSquare Roots on Both Sides


A) √(x + 1) = √3

(√(x + 1))2 = (√3)2

x + 1 = 3 x = 2

Square both sides.


Simplify.

√(x + 1) = √3 √(2 + 1) √3 √3 √3


CONFIDENTIAL 17

B) √(x + 8) - √(3x) = 0

√(x + 8) = √(3x)

(√(x + 8))2 = (√(3x))2

x + 8 = 3x 2x = 8

x = 4

Square both sides.

Check:

Add √(3x) from both sides.

Subtract x from both sides.


CONFIDENTIAL 18

Now you try!


CONFIDENTIAL 19

Squaring both sides of an equation may result in an extraneous solution —

a number that is not a solution of the original equation.

Suppose your original equation is x = 3.

Square both sides.

Now you have a new equation.

Solve this new equation for x by taking the square root of both sides.

Now there are two solutions. One (x = 3) is the original equation. The other (x = -3) is extraneous—it is not a

solution of the original equation. Because of extraneous solutions, it is important to check your answers.

x = 3

x2 = 9

√(x)2 = √9

x = 3 or x = -3

CONFIDENTIAL 20


Solve √(6 – x) = x. Check your answer.

(√(6 – x))2 = (x)2

6 – x = x2

x2 + x - 6 = 0

(x - 2) (x + 3) = 0

x - 2 = 0 or x + 3 = 0

x = 2 or x = -3

Square both sides.

Write in standard form.

Factor.

Zero-Product Property

Solve for x.

CONFIDENTIAL 21

Check:

Substitute 2 for x in the equation.

Substitute -3 for x in the equation.

-3 does not check; it is extraneous. The only solution is 2.

CONFIDENTIAL 22

Now you try!

Multiply. Write each product in simplest form.

CONFIDENTIAL 23

Geometry ApplicationGeometry Application

A rectangle has an area of 52 square feet. Its length is 13 feet, and its width

is √x feet. What is the value of x? What is the width of the rectangle?

Use the formula for area of a rectangle.

Substitute 52 for A, 13 for l, and √ x for w.


Square both sides.

CONFIDENTIAL 24

Check:

Substitute 16 for x in the equation.

The value of x is 16. The width of the rectangle is √(16) = 4 feet.

CONFIDENTIAL 25

6) A rectangle has an area of 15 cm2 . Its width is 5 cm, and its length is √(x + 1) cm. What is the value

of x? What is the length of the rectangle?

Now you try!

CONFIDENTIAL 26

Assessment

1) Is x = √3 a radical equation? Why or why not?

CONFIDENTIAL 27


CONFIDENTIAL 28


6)

7)

CONFIDENTIAL 29


8)

9)

CONFIDENTIAL 30

10) A trapezoid has an area of 14 cm2 . The length of one base is 4 cm and the length of the other

base is 10 cm. The height is √(2x + 3) cm. What is the value of x? What is the height of the

trapezoid?

CONFIDENTIAL 31

Solving Radical EquationsSolving Radical Equations

A radical equation is an equation thatcontains a variable within a radical. Inthis course, you will only study radicalequations that contain square roots.

Recall that you use inverse operations to solve equations. For nonnegative numbers,

squaring and taking the square root are inverse operations. When an equation

contains a variable within a square root, square both sides of the equation to solve.

Let’s review

CONFIDENTIAL 32

Power Property of Power Property of EqualityEquality

WORDS NUMBERS ALGEBRA

You can square both sides of

an equation, and the resulting

equation is still true.

3 = 1 + 2

(3)2 + (1 + 2)2

9 = 9

If a and b are real numbers

and a = b,

then a2 = b2.

CONFIDENTIAL 33

Solving Radical Equations by Adding or SubtractingSolving Radical Equations by Adding or Subtracting


A) √x + 3 = 10

√x = 7

√(x)2 = 72

x = 49


Square both sides.

Check:


Simplify.

√x + 3 = 10 √(49) + 3 10 7 + 3 10 10 10

CONFIDENTIAL 34

Solving Radical Equations by Multiplying or Solving Radical Equations by Multiplying or DividingDividingSolve each equation. Check your answer.

A) 3√x = 21

Method 1:√x = 7√(x)2 = 72

x = 49

Method 2:3√x = 21(3√(x))2 = (21)2

9x = 441 x = 49


Square both sides.


Simplify.

3√x = 21 3√(49) 21 3(7) 21 21 21


Square both sides.

CONFIDENTIAL 35



A) √(x + 1) = √3

(√(x + 1))2 = (√3)2

x + 1 = 3 x = 2

Square both sides.


Simplify.

√(x + 1) = √3 √(2 + 1) √3 √3 √3


CONFIDENTIAL 36

Squaring both sides of an equation may result in an extraneous solution —

a number that is not a solution of the original equation.

Suppose your original equation is x = 3.

Square both sides.

Now you have a new equation.

Solve this new equation for x by taking the square root of both sides.

Now there are two solutions. One (x = 3) is the original equation. The other (x = -3) is extraneous—it is not a

solution of the original equation. Because of extraneous solutions, it is important to check your answers.

x = 3

x2 = 9

√(x)2 = √9

x = 3 or x = -3

CONFIDENTIAL 37


Solve √(6 – x) = x. Check your answer.

(√(6 – x))2 = (x)2

6 – x = x2

x2 + x - 6 = 0

(x - 2) (x + 3) = 0

x - 2 = 0 or x + 3 = 0

x = 2 or x = -3

Square both sides.

Write in standard form.

Factor.

Zero-Product Property

Solve for x.

CONFIDENTIAL 38

Geometry ApplicationGeometry Application

A rectangle has an area of 52 square feet. Its length is 13 feet, and its width

is √x feet. What is the value of x? What is the width of the rectangle?

Use the formula for area of a rectangle.

Substitute 52 for A, 13 for l, and √ x for w.


Square both sides.

CONFIDENTIAL 39

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Algebra1 Solving Radical Equations