§ 7.6

Radical Equations
A radical equation is an equation in which the variable occurs in a square root, cube root, or any higher root. In this section, you will learn how to solve radical equations.
When the variable occurs in a square root, it is necessary to square both sides of the equation. When you square both sides of an equation, sometimes extra answers creep in, called extraneous roots. For example, consider the following very simple original equation.
x x x
2

2



4
2
Square both sides.
Solve the equation. But -2 does not work in the original equation. Then -2 is an extraneous root. It’s like a hitchhiker that we picked up when we squared both sides. It works only in the squared form and is not a root of the original.
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 7.6

Radical Equations
Just to note then….
When you solve a rational equation and must raise both sides to an even power, remember to check your roots. Throw out any extra (extraneous) roots from your solution set.
Another thing that you should know is … if your equation contains two or more square root expressions, you will need to isolate one square root expression, square both sides, and then you may have to repeat the process. You may have to square both sides twice.
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 7.6

Solving Radical Equations
1) If necessary, arrange terms so that one radical is isolated on one side of the equation.
2) Raise both sides of the equation to the n th power to eliminate the n th root.
3) Solve the resulting equation. If this equation still contains radicals, repeat steps 1 and 2.
4) Check all proposed solutions in the original equation .
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 7.6

Solving Radical Equations
Check Point 1
Solve: 3 x

4

8 .
SOLUTION
1) Isolate a radical on one side.
3 x

4

8 .
2) Raise both sides to the nth power.
Because n , the index, is
2, we square both sides.

3 x

4
2


 
2
3 x

4

64 Simplify.
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 7.6

Solving Radical Equations
CONTINUED
3) Solve the resulting equation.
3 x

4

64
3 x
 x

60
20
This is the equation from step 2.
Subtract 5 from both sides.
Divide both sides by 2.
4) Check the proposed solution in the original equation.
Check 10:
3 x

4

8
3
 

4 8
64
 ?
8
8

8 true
The solution is 20.
60

4
 ?
8 The solution set is {20}.
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 7.6

Solving Radical Equations
Number 4
Solve: 3 x

2

5

0 .
SOLUTION
1) Isolate a radical on one side. The radical can be isolated by adding 5 to both sides. We obtain
3 x

2

5 .
2) Raise both sides to the nth power.
3 x

2

 
2
3 x

2

25
Simplify.
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 7.6

Solving Radical Equations
CONTINUED
3) Solve the resulting equation.
3 x
 x

27
9
3 x

2

25
4) Check the proposed solution in the original equation.
Check 9:
3 x

2

5

0
3
 

2 5

0
0

0 true
The solution is 9.
25

5

0
?
The solution set is {9}.
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 7.6

Solving Radical Equations
EXAMPLE (Also number 6)
Solve: 2 x

5

11

6 .
SOLUTION
1) Isolate a radical on one side.
The radical, , can be isolated by subtracting 11 from both sides. We obtain
2 x

5
 
5 .
2) Raise both sides to the nth power.
Because n , the index, is
2, we square both sides.

2 x

5
2


 
2
2 x

5

25 Simplify.
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 7.6

Solving Radical Equations
CONTINUED
3) Solve the resulting equation.
2 x

5

25
2 x
 x

20
10
This is the equation from step 2.
Subtract 5 from both sides.
Divide both sides by 2.
4) Check the proposed solution in the original equation.
Check 10:
2 x

5

11

6
2
 

5

11 6
25

11

6
5

11

6
16

6 false
Therefore there is no solution to the equation.
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 7.6

Solving Radical Equations
EXAMPLE
Solve: 3 x

3 x

7
 
5 .
SOLUTION
1) Isolate a radical on one side.
The radical, , can be isolated by subtracting 3 x from both sides. We obtain

3 x

7
 
5

3 x .
2) Raise both sides to the nth power.
Because n , the index, is
2, we square both sides.


3 x

7

2 


5

3 x

2
3 x

7

25

30 x

9 x
2
Simplify. Use the special formula

A

B

2 
A
2

2 AB

B
2
.
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 7.6

Solving Radical Equations
CONTINUED x
2
3) Solve the resulting equation.
Because of the -term, the resulting equation is a quadratic equation. We need to write this quadratic equation in standard form. We can obtain zero on the left side by subtracting 3 x and 7 from both sides.
3 x

7

9 x
2 
30 x

25
0

9 x
2 
27 x

18
0

9
 x
2 
3 x

2

0
0
 x
2 
3 x

2

 x

2
 x

1

Equation from step 2.
Subtract 3 x and 7 from both sides.
Factor out the GCF, 9.
Divide both sides by 9.
Factor the right side.
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 7.6

Solving Radical Equations
CONTINUED x

2
 x

0

2 x

1
 x

0

1
Set each factor equal to 0.
Solve for x .
4) Check the proposed solutions in the original equation.
Check -2:
3 x

3 x
3
 

3
 


7
7
 
5

5
Check -1:
3 x

3 x
3
 

3
 


7
7
 
5

5

6

1

5

6

1

5

7
 
5 false

3

4

5

3

2

5

5
 
5
The solution is -1. The solution set is {-1}.
true
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 7.6

Solving Radical Equations
Number 10
Solve:
SOLUTION
2 x

1
 x

7 .
x

12
Do on board
4) Check the proposed solution in the original equation.
Check 12:
2 x

1
 x

7
2
 

1 12

7
5

5 true
25

5
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 7.6

In Summary…
Sometimes proposed solutions will work in the final simplified form of the original equation, but will not work in the original equation itself. These imposter roots that sometimes slip in when we square both sides of an equation are called extraneous roots.
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 7.6

DONE

Solving Radical Equations
EXAMPLE
Solve: x

4
 x

4

4 .
SOLUTION
1) Isolate a radical on one side.
The radical, , can be x

4

4
 x

4 .
2) Raise both sides to the nth power.
Because n , the index, is

2, we square both sides.
x

4
2
 
4
 x

4
2
 x

4

16

8 x

4
 x

4
Simplify. Use the special formula

A

B

2 
A
2

2 AB

B
2
.
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 7.6

Solving Radical Equations
CONTINUED x

4

20
 x

8 x

4 Combine like terms.
1) Isolate a radical on one side.
The radical, can be isolated by subtracting 20 + x from both sides and then dividing both sides by -8. We obtain
3
 x

4 .
2) Raise both sides to the nth power.
Because n , the index, is
2, we square both sides.
3
2 
 x

4

2
9
 x

4 Square the 3 and the x

4 .
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 7.6

Solving Radical Equations
CONTINUED
3) Solve the resulting equation.
9
 x

4
This is the equation from the last step.
5
 x Subtract 4 from both sides.
3) Check the proposed solution in the original equation.
Check 5: x

4


4
 x

4

 

4 4
4
1

9 4
1

3

4
4

4 true
The solution is 5. The solution set is {5}.
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 7.6

Solving Radical Equations
EXAMPLE
Solve:

2 x

3
 1
4

7

10 .
SOLUTION
Although we can rewrite the equation in radical form
4 2 x

3

7

10 , it is not necessary to do so. Because, the equation involves a fourth root, we isolate the radical term – that is, the term with the rational exponent – and raise both sides to the 4 th power.

2 x

3
 1
4

7

10 This is the given equation.

2 x

3
 1
4

3 Subtract 7 from both sides.
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 7.6

Solving Radical Equations
CONTINUED



2 x

3
 1
4


4

 
4
2 x

3

81
2 x

78 x

39
Raise both sides to the 4 th power.
Multiply exponents on the left sides and then simplify.
Subtract 3 from both sides.
Divide both sides by 2.
Upon checking the proposed solution, 39, in the original equation, we find that it checks out and is a solution. Therefore the solution set is {39}.
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 7.6

Solving Radical Equations
EXAMPLE
For each planet in our solar system, its year is the time it takes the planet to revolve once around the sun. The function f
 

0 .
2 x
3
2 models the number of Earth days in a planet’s year, f ( x ), where x is the average distance of the planet from the sun, in millions of kilometers. Use the function to solve the following problem.
There are approximately 88 Earth days in the year of the planet
Mercury. What is the average distance of Mercury from the sun?
Use a calculator and round to the nearest million kilometers.
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 7.6

Solving Radical Equations
CONTINUED
SOLUTION
To find the average distance of Mercury from the sun, replace f ( x ) in the function with 88.
f
 

0 .
2 x
3
2
88

0 .
2 x 2
3
440
 x
3
2
440
2 
 
2
193 , 600
 x
3
This is the given equation.
Replace f ( x ) with 88.
Divide both sides by 0.2.
Square both sides.
Simplify.
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 7.6

Solving Radical Equations
CONTINUED
3 193 , 600
 3 3 x
58
 x
Take the cube root of both sides.
Simplify.
The model indicates that the average distance, to the nearest million kilometers, that Mercury is from the sun is 58 million kilometers.
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 7.6

DONE