Radical Equations

Grade 9 · algebra · 86 practice problems · read aloud

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Radical Equations: Unlocking the Mystery 🗝️

A radical equation is an equation where the variable is inside a radical symbol, like a square root. Solving them is like a puzzle—you need to "free" the variable to find its value. This skill is super useful in geometry (like finding a side of a right triangle) and real-world problems involving areas and distances.

How to Solve Radical Equations: A Step-by-Step Guide

  1. Isolate the Radical: Get the radical expression (like √x) by itself on one side of the equation.
  2. Square Both Sides: To cancel out a square root, square both sides of the equation. (If it were a cube root, you would cube both sides).
  3. Solve the New Equation: The radical is now gone! Solve the remaining equation for the variable.
  4. Check for Extraneous Solutions! 🔍 This is the most important step. Plug your answer(s) back into the original equation. Sometimes, the squaring step creates solutions that don't actually work.

Worked Examples

Example 1: Solve √(x + 2) = 5

  1. The radical is already isolated.
  2. Square both sides: (√(x + 2))² = (5)² → x + 2 = 25
  3. Solve for x: x = 23
  4. Check: √(23 + 2) = √25 = 5. It works! ✅

Solution: x = 23

Example 2: Solve √(2x - 3) + 4 = 9

  1. Isolate the radical: √(2x - 3) = 5
  2. Square both sides: 2x - 3 = 25
  3. Solve for x: 2x = 28 → x = 14
  4. Check: √(2(14) - 3) + 4 = √(25) + 4 = 5 + 4 = 9. It works! ✅

Solution: x = 14

Common Mistakes to Avoid

Forgetting to Check for Extraneous Solutions: This is the #1 mistake! Always verify your answer in the original equation.

Squaring Incorrectly: Remember, (a + b)² is NOT a² + b². You must square the entire side. For example, (√x + 2)² means (√x + 2)(√x + 2).

Not Isolating First: If you have other terms with the radical, isolate it before squaring. Squaring too early makes a big mess!

Tips & Tricks

Memory Aid: Think "Isolate, Annihilate, Evaluate, Validate." You isolate the radical, annihilate it by squaring, evaluate for x, and validate your answer.

Domain Check: Before you start, remember the expression under a square root must be ≥ 0. This can help you spot impossible answers early.

How to Practice

  • Start with simple equations where the radical is already isolated.
  • Move on to problems where you need to add, subtract, or multiply first.
  • Create your own practice problems by starting with a number, say x=6, and building an equation from it (e.g., √(6-2)=2). Then solve it to see if you get 6!
  • Always write out the check step until it becomes a habit.

Practice problems

6 of the 86, worked through step by step — try them before opening the answer.

1 √(2x + 3) = 5

Hint: To solve radical equations, isolate the radical expression and then square both sides to eliminate the square root. Remember to check your solution in the original equation.

Show the answer

Answer: 11

  1. Recognize that the square root is already isolated on the left side.
  2. To remove the square root, square both sides of the equation. This gives: (sqrt(2x + 3))^2 = 5^2
  3. Simplify both sides. On the left: squaring the square root leaves just the expression inside: 2x + 3. On the right: 5 squared is 25. So we have: 2x + 3 = 25
  4. Subtract 3 from both sides to isolate the term with x. 2x + 3 - 3 = 25 - 3 2x = 22
  5. Divide both sides by 2 to solve for x. 2x / 2 = 22 / 2 x = 11
  6. Check the solution in the original equation to ensure it is valid. Substitute x = 11 into sqrt(2x + 3): sqrt(2*11 + 3) = sqrt(22 + 3) = sqrt(25) = 5 This matches the right-hand side of the original equation, so the solution is correct. Final answer: x = 11

We are solving the equation: sqrt(2x + 3) = 5

2 √(3x + 7) = 4

Hint: To solve radical equations, first isolate the radical expression, then square both sides to eliminate the square root. Remember to check your solution in the original equation.

Show the answer

Answer: 3

  1. Since the square root is already isolated, square both sides to remove the square root. (√(3x + 7))^2 = 4^2
  2. Simplify both sides. Left side: (√(3x + 7))^2 = 3x + 7 Right side: 4^2 = 16 So we have: 3x + 7 = 16
  3. Subtract 7 from both sides to isolate the term with x. 3x + 7 - 7 = 16 - 7 3x = 9
  4. Divide both sides by 3 to solve for x. 3x / 3 = 9 / 3 x = 3
  5. Check the solution in the original equation. Original: √(3x + 7) = 4 Substitute x = 3: √(3*3 + 7) = √(9 + 7) = √16 = 4 This matches the right-hand side, so the solution is correct. Final answer: x = 3

We are solving the equation: √(3x + 7) = 4

3 √(3x + 7) = 5

Hint: To solve radical equations, isolate the radical expression and then square both sides to eliminate the square root. Remember to check your solution in the original equation.

Show the answer

Answer: 6

  1. Understand the equation** The square root of (3x + 7) equals 5. Since the square root symbol √ means the principal (non-negative) square root, both sides are non-negative. **
  2. Eliminate the square root** To remove the square root, square both sides of the equation: [ √(3x + 7) ]^2 = (5)^2 This simplifies to: 3x + 7 = 25 **
  3. Isolate the term with x** Subtract 7 from both sides: 3x + 7 - 7 = 25 - 7 3x = 18 **
  4. Solve for x** Divide both sides by 3: x = 18 / 3 x = 6 **
  5. Check the solution** Substitute x = 6 into the original equation: √(3*6 + 7) = √(18 + 7) = √25 = 5 This matches the right-hand side, so the solution is correct. **Final answer:** x = 6

We are solving the equation: √(3x + 7) = 5 **

4 √(3x + 7) - 2 = 5

Hint: First isolate the radical term by moving the constant to the other side, then square both sides to eliminate the square root. Remember to check your solution in the original equation.

Show the answer

Answer: 14

  1. Isolate the square root term. Add 2 to both sides: √(3x + 7) = 5 + 2 √(3x + 7) = 7
  2. Square both sides to remove the square root. (√(3x + 7))^2 = 7^2 3x + 7 = 49
  3. Solve for x. Subtract 7 from both sides: 3x = 49 - 7 3x = 42 Divide both sides by 3: x = 42 / 3 x = 14
  4. Check the solution in the original equation. √(3*14 + 7) - 2 = √(42 + 7) - 2 = √49 - 2 = 7 - 2 = 5 This matches the original equation, so the solution is correct. Final answer: x = 14

We start with the equation: √(3x + 7) - 2 = 5

5 √(2x + 5) - 3 = 0

Hint: Isolate the radical expression first, then square both sides to eliminate the square root. Remember to check your solution in the original equation.

Show the answer

Answer: 2

  1. Isolate the square root term. Add 3 to both sides of the equation: √(2x + 5) = 3
  2. Square both sides to remove the square root. (√(2x + 5))^2 = (3)^2 This simplifies to: 2x + 5 = 9
  3. Solve for x. Subtract 5 from both sides: 2x = 4 Divide both sides by 2: x = 2
  4. Check for extraneous solutions by substituting back into the original equation. Original equation: √(2x + 5) - 3 = 0 Substitute x = 2: √(2*2 + 5) - 3 = √(4 + 5) - 3 = √9 - 3 = 3 - 3 = 0 This is correct, so x = 2 is valid. Final answer: 2

We start with the equation: √(2x + 5) - 3 = 0

6 √(4x - 7) + 2 = 9

Hint: Isolate the radical term first before squaring both sides. Remember to check for extraneous solutions.

Show the answer

Answer: 14

  1. Isolate the radical term √(4x - 7) + 2 = 9 √(4x - 7) = 9 - 2 √(4x - 7) = 7
  2. Square both sides to eliminate the radical (√(4x - 7))² = 7² 4x - 7 = 49
  3. Solve for x 4x = 49 + 7 4x = 56 x = 56 ÷ 4 x = 14
  4. Check for extraneous solutions Substitute x = 14 back into the original equation: √(4(14) - 7) + 2 = √(56 - 7) + 2 = √49 + 2 = 7 + 2 = 9 The solution checks out.

The answer is 14.

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