Intermediate Algebra – 18
HFCC Math Lab
SOLVING RADICAL EQUATIONS
You already know how to solve linear equations and quadratic equations by factoring. In this handout, you are
going to learn how to solve radical equations (1) containing one radical expression and (2) containing two
radical expressions. But first let us answer the question: “What is a radical equation”? The answer is: a radical
equation is an equation in which the variable we are solving for appears inside the radical symbol.
For example,
2 x 1 3,
3
5 y 2 5 3, and x 7
x 2
5 are all radical equations. Let us now
proceed to solve radical equations.
Part 1 Radical Equations Containing One Radical Expression
Follow the following steps to solve such radical equations.
Step I.
Isolate the radical expression by itself on one side of the
equation.
Step II.
Get rid of the radical by raising both sides to the
power
where n denotes the index of the radical. In simple English, it
means that we square both sides if the radical is the principal
square root, we cube both sides if the radical is the cube root,
we raise both sides to the power 4 if the radical is the principal
fourth root, and so on.
Step III.
Simplify both sides of the equation in Step II and solve the
resulting equation.
Step IV.
Check the values obtained in Step III if the index of the radical
in the original equation is even. There is no need to check the
values obtained in Step III if the index of the radical in the
original equation is odd.
Step V.
The solutions to the original equation are those values of the
variable from Step IV which checked into the original equation.
Revised 11/09
1
Please read the following examples very carefully.
Examples:
1.
Solution:
Solve each of the following radical equations
5
3 2x
2
To isolate the radical, we add 5 to both sides.
5
3 2x
2+ 5
3 2x = 3
Square both sides.
2
3 2x
3
2
3 – 2x = 9
Solve this linear equation for x.
Subtract 3 from both sides.
3 2x 3 9 3
2x
6
2x
2
6
2
So, x
3
Check
5
3 2x
2
Replace x by -3
3 2( 3)
5
2
9 5
2
3 5
2
2
2
Therefore, the solution is x =
Revised 11/09
3
2
3 2x 2
2.
5
Proceed exactly like the first example.
Add 2 to both sides.
3 2x
5 2
3 2x
3
Since the principal square root (in fact, the principal even root) of any expression cannot be negative, the
original equation has no solution.
NOTE:
If you had continued after isolating the radical, you will get x = 3 .
But then x = 3 will not satisfy the original equation. So, once again,
the original equation has no solution.
NOTE:
Do you understand everything written in this handout so far? If your answer is “no”, please read
again and then proceed further.
3.
3
2
3 4x
Solution:
3
To isolate the radical, we subtract 2 from both sides.
3
3 4x = 3 – 2
3
3 4x = 5
Cube both sides.
3
3
3 4x
5
3 4x
3
125
Subtract 3 from both sides
4x
128
128
x
4
x 32
NOTE:
32
Since the equation involved the cube root, there is no need to check the solution.
4.
2x 5
4 x
Subtract 4 from both sides.
Revised 11/09
3
2x 5
x 4
Square both sides
2
2x 5
x 4
2x 5
2
x 4 x 4
2 x 5 x 2 8 x 16
This is a quadratic equation.
Interchange the two sides.
x2
8 x 16 2 x 5
Get zero on the right side.
x2
8 x 16 2 x 5 0
x 2 10 x 21 0
Factor the left side.
x 3 x 7
So,
0
x 3 0 or x 7 0
x 3 or x 7
Check for x = 3
2x 5
Check for x = 7
2x 5
4 x
Replace x by 3. We get
4 x
Replace x by 7. We get
2 3 5 4 3
2 7 5 4 7
1 4 3
1 4 3
9 4 7
3 4 7
5 3 false
7 7 true
Since x = 3 does not check and x = 7 checks, therefore the only solution to the original equation is x = 7.
Note: x = 3 called an extraneous solution
Revised 11/09
4
Part II
Radical equations containing two radical expressions when both the radicals are
principal square roots.
Follow the following steps to solve such radical equations.
STEP I
Write the equation so that one radical expression (involving the
principal square root) appears on each side of the equation.
STEP II
Square both sides of the equation very carefully.
STEP III
Solve the equation obtained in Step II. If one radical expression
(involving the principal square root) still remains, solve this equation
as we did in Part I of this handout.
STEP IV
Check the values obtained in Step III in the original equation.
STEP V
The solutions to the original equation are those values of the variable
from Step IV which checked into the original equation.
Please read the following examples very carefully.
Examples:
Solve each of the following radical equations.
1.
Solution:
3p 4
2 5p 0
First move one of the two radicals to the right side of the equation. So, we add
to both sides and obtain
3p 4
2 5p
Square both sides
2
2
3p 4
2 5p
3p 4 2 5p
Solve this linear equation
3p
p
Check:
5p 2 4
8p 6
6 3
8 4
3p 4
2 5p 0
Replace p by
Revised 11/09
3
we get
4
5
2 5p
3
3
4
4
2 5
3
4
2
15
4
9
4
4
7
4
0
0
7
0
4
Since the principal square root of a negative number is not a real number, therefore the equation has no
solution.
2.
4 3p
5p 2
0
Proceed just like the first example and get p
Check:
4 3p
5p 2
Replace p by
3
.
4
0
3
we get
4
4 3
4
3
4
3
2
4
0
15
2
4
0
7
4
7
4
0
7
2
7
0
2
0 0 true
5
9
4
The solution to the equation is p
x 3
3.
x 4
3
4
1
First move one of the two radicals to the right side of the equation. So, we
add
x 4 to both sides and get
x 3
Revised 11/09
x 4
1
6
Square both sides
2
2
x 3
x 4 1
Be careful when you square on the right side.
x 3
x 4 1
x 4 1
Use FOIL on the right side.
x 3 x 4
x 4
x 4 1
x 3 x 5 2 x 4
We are now left with one radical expression. Isolate this radical expression
on one side. Subtract x and subtract 5 from both sides.
x 3 x 5
2 x 4
8
2 x 4
Divide both sides by -2.
8
2
2 x 4
2
4
x 4
Square both sides to get
16 x 4
So, x 12
Check:
x 3
12 3
x 4
1
12 4
1
9
16
1
3 4 1
1 1(true)
The solution is x = 12
Revised 11/09
7
EXERCISES
Solve each of the following radical equations.
1.
4x 3 7
2
2.
5x 4 3
1
3.
4x 3 2
7
4.
5x 4 1
3
x2 2 x 4 5
5. 3
7.
3
4x 1 2 0
9. y
x2 3x 6
6. 5
8.
5y 4
3
8 5x 3 2
10. p
6 p
11.
69 6 x 7
x
12.
3 x 10 3 x
13.
69 6 x x 7
14.
x 4 3x
2 x 2 3x 8 6 7
16.
17.
5x 1
3x 1
18.
y 5
19.
2n 3
n 2
2
20.
y 4
2p 9
22.
2y 3
24.
x 2
15.
4
21. 2
23.
Revised 11/09
p 1
1 2y
y 5
0
8
4
7
4
2 x 2 3x 8 7 6
5
y
y 7
3
4y 1 0
5x 1 3
ANSWERS AND SOLUTIONS
1.
4x 3 7
2
Add 7 to both sides.
4x 3 5
Square both sides to get
4x 3=25
4 x 28
x 7
Check:
4x 3 7
2
4 7 3 7
2
25 7
2
5 7
2
2
2 true
The solution is x 7
2. x
3.
8
5
4x 3 2
7
Add 2 to both sides.
4x 3
5
Since the principal square root of a number cannot be negative, the equation has no solution.
4.
No solution.
Revised 11/09
9
x2 2 x 4
5. 3
5
Subtract 3 from both sides.
x2 2x 4 2
Square both sides. We get
x2
2x 4
4
This is a quadratic equation.
x2
2x 4 4 0
x2
2x 8 0
Factor to get x 2 x 4
0
x 2 0 or x 4 0
x
2 or x
4
Check for x
Check for x
x2 2 x 4
3
3
2
2
2
2
3
2
5
4
4
4 4 4
5
3
4
3
3
5
4
x2 2 x 4
5
42 2 4 4
5
3
5
4
3 2 5
3 2 5
5 5(True)
5=5(True)
The two solutions are x =
6. x
7.
2, x 5
3
4x 1 2
0
Subtract 2 from both sides.
3
4x 1
2
Cube both sides.
3
3
4x 1
2
4x 1
8
4x
7
x
7
4
3
The solution is x
Revised 11/09
7
4
10
2, 4.
9
5
8. x
9. y
5y 4
Square both sides.
y2
5y 4
y2 5 y 4 0
y 1 y 4
0
y 1 0 or y 4 0
y 1or y
4
Show yourself that both these values check. Therefore,the two solutions
are y 1, y
10. p
11.
4
2is the only solution (p
3does not check).
69 6 x 7 x
Subtract 7 from both sides.
69 6 x x 7
Square both sides.
2
69 6 x
69 6 x
x 7
2
x 2 14 x 49
x 2 14 x 49 69 6 x
x 2 14 x 49 69 6 x
0
x 2 8 x 20 0
x 2 x 10
0
x 2 0 or x 10 0
x
2 or x 10
Show yourself that x = 2 does not check and x = 10 checks. So, the only solution to the
equation is x = 10.
12. x
Revised 11/09
1
is the only solution x
3
2does not check
11
13.
69 6 x x 7
Subtract x from both sides.
69 6 x 7 x
.
Proceed exactly like problem 11. You will get x
2 or x 10
Show yourself that x
2 checks and x 10 does not check. So,
the only solution to the equation is x
2.
16
is the only solution x 1does not check .
9
14. x
15.
2 x 2 3x 8 6 7
Subtract 6 from both sides.
4
2 x 2 3x 8 1
Raise both sides to the power 4. We get
4
4
2 x 2 3x 8
4
1
4
2x 2 3 x 8 1
2 x 2 3x 8 1 0
2 x 2 3x 9 0
2x 3 x 3
0
2 x 3 0 or x 3 0
3
or x 3
2
Show yourself that both these values check. So, the two solutions are
x
x
3
,x 3
2
16. No solution.
Revised 11/09
12
17.
5x 1
3x 1
Add 3 x to both sides
5x 1
3x 1
Square both sides
2
2
5x+1
3x 1
5x 1
3x 1
3x 1
3x
3x 1
5 x 1 3x
5 x 1 3x 2 3x 1
5 x 1 3x 1 2 3x
2 x 2 3x
Divide both sides by 2
x
3x
Square both sides
x 2 3x
This is a quadratic equation
x 2 3x 0
Factor the left side
x x 3
0
x
0 or x 3 0
x 0 or x 3
Show yourself that both these values check. Thus, the two solutions are
x 0, x 3
18. y
Revised 11/09
4
13
19.
2n 3
n 2
2
Add n 2 to both sides.
2n+3
n 2 2
Square both sides.
2
2
2n 3
n 2 2
2n 3
n 2 2
n 2 2
2n 3 n 2 2 n 2 2 n 2 4
2n 3 n 2 4 n 2
Subtract n and subtract 2 from both sides.
2n 3 n 2 4 n 2
n 1 4 n 2
Square both sides
n 1
2
n 1 n 1
2
4 n 2
42
2
n 2
n 2 2n 1 16 n 2
n 2 2n 1 16n 32
This is a quadratic equation.
Write it in standard form.
n 2 2n 1 16n 32 0
n 2 14n 33 0
n 3 n 11
0
n 3 0 or n 11 0
n 3 or n 11
Show yourself that both these values check.
Thus, the two solutions are n 3, n 11.
20. y
Revised 11/09
3
14
21. 2
p 1
2p 9
Square both sides.
2
2
2
p 1 2
2p 9
p 1
4 2 p 1 2 p 1
5
2
p 1
2p 9
p 1 2p 9
p 4 p 1
2p 9
Subtract 5 and subtract p from both sides.
We get
4 p 1
2p 9 5 p
4 p 1
p 4
Square both sides
2
4 p 1
p 4
2
16 p 1
p 2 8 p 16
16 p 16
p 2 8 p 16
p 2 8 p 16 16 p 16
p 2 8 p 16 16 p 16 0
p2 8 p
0
p p 8
0
So, p 0 or p 8 0
Show yourself that both these values check.
Therefore, the two solutions are p
22. y
Revised 11/09
0, p 8
2
15
23.
1 2y
Add
y 5
0
y+5 to both sides
1-2y
y 5
Square both sides
2
2
1 2y
y 5
1 2y
y 5
1 5
y 2y
4 3y
4
3
4
Show that y
checks
3
4
The solution is y
3
So, y
24. x
9
,x
4
3
NOTE: You can get additional instruction and practice by going to the following
web sites:
http://tutorial.math.lamar.edu/classes/Alg/SolveRadicalEqns.aspx
This website gives an excellent detailed example on solving radical equations.
http://www.purplemath.com/modules/solverad.htm
This website has some very good examples on solving radical equations with two radicals
with some graphical illustration.
Revised 11/09
16