Section 10.6 Radical Equations
Radical Equations
 A Radical Equation is an equation the variable occurs in radical(s).
Solving Radical Equations
 Step 1: Isolate the radical on one side of the equation, if necessary.
 Step 2: Eliminate the radical by raising both sides of the equation to nth power.
 Step 3: Repeat the step 1 and the step 2 if still radical(s) remain(s).
 Step 4: Solve for the variable.
 Step 5: Check all proposed solutions in the original equation
Exercises
(Solution 1)
Step 1: Isolate the radical, if necessary. Skip!
Step 2: Raise both sides to nth power.
Since the index is 2, raise both sides to 2nd power.
√3𝑥 − 5 = 4
2
2
�√3𝑥 − 5� = 4
3𝑥 − 5 = 16
Step 3: Repeat the step 1 and 2. No more radicals. Skip!
Step 4: Solve
Step 5: Check
Since the proposed solution
3𝑥 − 5 = 16
is 7, substitute 7 for x.
+5 +5
3𝑥 = 21
√3𝑥 − 5 = 4
3𝑥 21
�3(7) − 5 = 4
=
3
3
√21 − 5 = 4
𝑥= 7
√16 = 4
𝑻𝒓𝒖𝒆 ⟹
4= 4
The statement is true for x =7. Thus, the solution is 7
(Solution 2)
Step 1: Isolate the radical, if necessary.
√3𝑥 − 5 − 7 = 0
+7 +7
√3𝑥 − 5 = 7
Step 2: Raise both sides to nth power.
Since the index is 2, raise both sides to 2nd power.
√3𝑥 − 5 = 7
2
2
�√3𝑥 − 5� = 7 ⟹ 3𝑥 − 5 = 49
Step 3: Repeat the step 1 and 2. No more radicals. Skip!
Step 4: Solve
Step 5: Check
Since the proposed solution
3𝑥 − 5 = 49
is 18, substitute 18 for x.
+5 +5
3𝑥 = 54
√3𝑥 − 5 − 7 = 0
3𝑥 54
�3(18) − 5 − 7 = 0
=
3
3
√54 − 5 − 7 = 0
𝑥 = 18
√49 − 7 = 0
7 − 7= 0
𝑻𝒓𝒖𝒆 ⟹
0= 0
The statement is true for x = 18, thus the solution is 18.
Cheon-Sig Lee
www.coastalbend.edu/lee
Page 1
Section 10.6 Radical Equations
(Solution 3)
Step 1: Isolate the radical, if necessary.
√2𝑥 + 3 + 5 = 2
−5 −5
√2𝑥 + 3 = −3
√2𝑥 + 3 is a principal square root and the principal square
root cannot be negative. Therefore, there is no solution.
(Proof)
The proposed solution is 3.
√2𝑥 + 3 = −3
2
2
(−3)
=
√2𝑥 + 3 + 5 = 2
�√2𝑥 + 3�
2𝑥 + 3 = 9
�2(3) + 3 + 5 = 2
2𝑥 = 6
√9 + 5 = 2
𝑥= 3
𝑭𝒂𝒍𝒔𝒆 ⟹
8= 2
Since 8 ≠ 2, false for 3. Thus, there is no solution.
(Solution 4)
Step 1: Isolate the radical. Skip!
Step 2: Raise both sides to nth power.
Since the index is 2, raise both sides to 2nd power.
𝑥 = √2𝑥 + 15
2
𝑥 2 = �√2𝑥 + 15�
⟹ 𝑥 2 = 2𝑥 + 15
Step 3: Repeat the step 1 and 2. No more radical. Skip!
Step 4: Solve for the variable
𝑥 2 = 2𝑥 + 15
−2𝑥 − 15 −2𝑥 − 15
𝑥 2 − 2𝑥 − 15 = 0
By ac-method, we have
(𝑥 + 3)(𝑥 − 5) = 0
or
𝑥 + 3= 0
𝑥 − 5= 0
−3 −3
+5 +5
𝑥 = −3
𝑥= 5
Proposed solutions are −3, 5
Step 5: Check
Check 𝑥 = −3
Check 𝑥 = 5
𝑥 = √2𝑥 + 15
𝑥 = √2𝑥 + 15
−3 = �2(−3) + 15
5 = �2(5) + 15
−3 = √−6 + 15
5 = √10 + 15
−3 = √9
5 = √25
−3 = 3 ⟸ 𝐹𝑎𝑙𝑠𝑒
5 = 5 ⟸ 𝑇𝑟𝑢𝑒
‘False’ for −3 and ‘True’ for 5.
Therefore, the solution is only 5.
Cheon-Sig Lee
www.coastalbend.edu/lee
Page 2
Section 10.6 Radical Equations
(Solution 5)
Step 1: Isolate the radical. Skip!
Step 2: Raise both sides to nth power.
Since the index is 2, raise both sides to 2nd power.
√13𝑥 + 25 = 𝑥 + 5
2
2
�√13𝑥 + 25� = (𝑥 + 5)
2
13𝑥 + 25 = 𝑥 + 10𝑥 + 25
Step 3: Repeat step 1 and 2. No more radical. Skip!
Step 4: Solve for the variable
13𝑥 + 25 = 𝑥 2 + 10𝑥 + 25
−13𝑥 − 25
−13𝑥 − 25
0 = 𝑥 2 − 3𝑥
By factoring out GCF, we have
0 = 𝑥(𝑥 − 3)
or
𝑥= 0
𝑥 − 3= 0
𝑥= 3
Proposed solutions are 0, 3
Step 5: Check
Check 𝑥 = 0
Check 𝑥 = 3
√13𝑥 + 25 = 𝑥 + 5
√13𝑥 + 25 = 𝑥 + 5
�13(0) + 25 = 0 + 5
�13(3) + 25 = 3 + 5
√25 = 5
√39 + 25 = 8
𝑇𝑟𝑢𝑒 ⟹ 5 = 5
√64 = 8
𝑇𝑟𝑢𝑒 ⟹ 8 = 8
True for both 0 and 3. Thus, the solution set is {0, 3}
(Solution 6)
Step 1: Isolate the radical.
𝑥 = √4𝑥 − 7 + 1
−1
−1
𝑥 − 1 = √4𝑥 − 7
Cheon-Sig Lee
Step 2: Raise both sides to nth power.
Since the index is 2, raise both sides to 2nd power.
2
(𝑥 − 1)2 = �√4𝑥 − 7�
𝑥 2 − 2𝑥 + 1 = 4𝑥 − 7
Step 3: Repeat step 1 and 2. No more radical. Skip!
Step 4: Solve for the variable
𝑥 2 − 2𝑥 + 1 = 4𝑥 − 7
−4𝑥 + 7 −4𝑥 + 7
𝑥 2 − 6𝑥 + 8 = 0
By ac-method, we get
(𝑥 − 2)(𝑥 − 4) = 0
or
𝑥 − 2= 0
𝑥 − 4= 0
𝑥= 2
𝑥= 4
Proposed solutions are 2, 4
Step 5: Check
Check 𝑥 = 2:
Check 𝑥 = 4
𝑥 = √4𝑥 − 7 + 1
𝑥 = √4𝑥 − 7 + 1
2 = �4(2) − 7 + 1
4 = �4(4) − 7 + 1
2 = √8 − 7 + 1
4 = √16 − 7 + 1
2 = √1 + 1
4 = √9 + 1
2 = 2 ⟸ 𝑇𝑟𝑢𝑒
4 = 4 ⟸ 𝑇𝑟𝑢𝑒
True for both 2 and 4. Thus, solutions are 2, 4
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Section 10.6 Radical Equations
(Solution 7)
Step 1: Isolate radicals. Skip!
Step 2: Raise both sides to nth power.
√10𝑥 − 2 = √9𝑥 + 1
2
2
�√10𝑥 − 2� = �√9𝑥 + 1�
10𝑥 − 2 = 9𝑥 + 1
Step 3: Repeat step 1 and 2. No more radical. Skip!
Step 4: Solve for the variable
10𝑥 − 2 = 9𝑥 + 1
−9𝑥
−9𝑥
𝑥 − 2= 1
+2 +2
𝑥= 3
Proposed solution is 3.
Step 5: Check
√10𝑥 − 2 = √9𝑥 + 1
�10(3) − 2 = �9(3) + 1
√30 − 2 = √27 + 1
⟸ 𝑇𝑟𝑢𝑒
√28 = √28
True for 3. Thus, the solution is 3.
Additional Problems
A.1. Solve the radical equation.
√2𝑥 − 3 − √𝑥 − 2 − 2 = −1
A.2. Solve the radical equation.
3
√6𝑥 − 3 = −3
A.3. Solve the radical equation.
3
3
√3𝑥 + 5 = √2𝑥 − 5
A.4. Solve the radical equation
3
√2𝑥 = √4𝑥
Cheon-Sig Lee
www.coastalbend.edu/lee
Page 4