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Solving Radical Equations

Equations containing radicals

Written and reviewed by the Study Mondo Education TeamLast updated
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Solving Radical Equations

Strategy

  1. Isolate the radical on one side
  2. Raise both sides to the appropriate power
  3. Solve the resulting equation
  4. Check for extraneous solutions

Why Check?

Squaring both sides can introduce extraneous solutions.

Always substitute back into the original equation!

Multiple Radicals

If there are two radicals:

  1. Isolate one radical
  2. Square both sides
  3. Isolate the remaining radical
  4. Square again
  5. Solve and check

Domain Restrictions

For x\sqrt{x}: must have x≥0x \geq 0

For x3\sqrt[3]{x}: xx can be any real number

Example

Solve: 2x+3=5\sqrt{2x + 3} = 5

Square both sides: 2x+3=252x + 3 = 25 2x=222x = 22 x=11x = 11

Check: 2(11)+3=25=5\sqrt{2(11) + 3} = \sqrt{25} = 5 ✓

📚 Practice Problems

1Problem 1easy

❓ Question:

Solve: √x = 5

💡 Show Solution

Step 1: Square both sides: (√x)² = 5² x = 25

Step 2: Check the solution: √25 = 5 ✓

Answer: x = 25

2Problem 2easy

❓ Question:

Solve: x+5=4\sqrt{x + 5} = 4

💡 Show Solution

Square both sides: x+5=16x + 5 = 16

Solve for xx: x=11x = 11

Check: 11+5=16=4\sqrt{11 + 5} = \sqrt{16} = 4 ✓

Answer: x=11x = 11

3Problem 3easy

❓ Question:

Solve: √(x + 3) = 4

💡 Show Solution

Step 1: Square both sides: (√(x + 3))² = 4² x + 3 = 16

Step 2: Solve for x: x = 16 - 3 x = 13

Step 3: Check the solution: √(13 + 3) = √16 = 4 ✓

Answer: x = 13

4Problem 4medium

❓ Question:

Solve: 3x−2+4=10\sqrt{3x - 2} + 4 = 10

💡 Show Solution

Step 1: Isolate the radical 3x−2=6\sqrt{3x - 2} = 6

Step 2: Square both sides 3x−2=363x - 2 = 36

Step 3: Solve 3x=383x = 38 x=383x = \frac{38}{3}

Check: 3(383)−2+4=38−2+4=36+4=6+4=10\sqrt{3(\frac{38}{3}) - 2} + 4 = \sqrt{38 - 2} + 4 = \sqrt{36} + 4 = 6 + 4 = 10 ✓

Answer: x=383x = \frac{38}{3}

5Problem 5medium

❓ Question:

Solve: √(2x - 1) = x - 2

💡 Show Solution

Step 1: Square both sides: (√(2x - 1))² = (x - 2)² 2x - 1 = x² - 4x + 4

Step 2: Rearrange to standard form: 0 = x² - 4x - 2x + 4 + 1 0 = x² - 6x + 5

Step 3: Factor: 0 = (x - 5)(x - 1)

Step 4: Solve: x = 5 or x = 1

Step 5: Check x = 5: √(2(5) - 1) = √9 = 3 5 - 2 = 3 ✓

Step 6: Check x = 1: √(2(1) - 1) = √1 = 1 1 - 2 = -1 ✗ x = 1 is EXTRANEOUS

Answer: x = 5

6Problem 6medium

❓ Question:

Solve: √(x + 7) - √(x - 5) = 2

💡 Show Solution

Step 1: Isolate one radical: √(x + 7) = 2 + √(x - 5)

Step 2: Square both sides: x + 7 = (2 + √(x - 5))² x + 7 = 4 + 4√(x - 5) + (x - 5) x + 7 = x - 1 + 4√(x - 5)

Step 3: Simplify: x + 7 - x + 1 = 4√(x - 5) 8 = 4√(x - 5) 2 = √(x - 5)

Step 4: Square again: 4 = x - 5 x = 9

Step 5: Check x = 9: √(9 + 7) - √(9 - 5) = √16 - √4 = 4 - 2 = 2 ✓

Answer: x = 9

7Problem 7hard

❓ Question:

Solve: x+7=x−5\sqrt{x + 7} = x - 5

💡 Show Solution

Step 1: Square both sides x+7=(x−5)2x + 7 = (x - 5)^2 x+7=x2−10x+25x + 7 = x^2 - 10x + 25

Step 2: Rearrange to standard form 0=x2−11x+180 = x^2 - 11x + 18

Step 3: Factor 0=(x−9)(x−2)0 = (x - 9)(x - 2)

Step 4: Solve x=9 or x=2x = 9 \text{ or } x = 2

Step 5: Check both solutions

For x=9x = 9: 9+7=16=4\sqrt{9 + 7} = \sqrt{16} = 4 and 9−5=49 - 5 = 4 ✓

For x=2x = 2: 2+7=9=3\sqrt{2 + 7} = \sqrt{9} = 3 and 2−5=−32 - 5 = -3 ✗

Answer: x=9x = 9 only (x = 2 is extraneous)

8Problem 8hard

❓ Question:

Solve: ∛(x + 1) = 2

💡 Show Solution

Step 1: Cube both sides: (∛(x + 1))³ = 2³ x + 1 = 8

Step 2: Solve for x: x = 7

Step 3: Check: ∛(7 + 1) = ∛8 = 2 ✓

Step 4: Note about odd roots: Unlike even roots, odd roots can have negative radicands For example, ∛(-8) = -2

Answer: x = 7

Explain using:

⚠️ Common Mistakes: Solving Radical Equations

Avoid these 3 frequent errors

🌍 Real-World Applications: Solving Radical Equations

See how this math is used in the real world

📝 Worked Example: Solving a Quadratic by Factoring

Problem:

Solve x2−5x+6=0x^2 - 5x + 6 = 0.

2Factor the quadratic
3Set each factor equal to zero

📌 Related Topics in Radical Functions

❓ Frequently Asked Questions

What is Solving Radical Equations?▾
Equations containing radicals
How can I study Solving Radical Equations effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 8 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Solving Radical Equations study guide free?▾
Yes — all study notes, flashcards, and practice problems for Solving Radical Equations on Study Mondo are free to access. No account is needed.
What course covers Solving Radical Equations?▾
Solving Radical Equations is part of the Algebra 2 course on Study Mondo, specifically in the Radical Functions section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Solving Radical Equations?▾
Yes, this page includes 8 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.