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Solve Radical Equations Calculator

Solve radical equations step by step. Enter the coefficients for your equation type and get detailed solutions with extraneous solution checking.

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Key Tip: Always check your solutions!

Squaring both sides can introduce extraneous solutions that don't satisfy the original equation.

√(ax + b) = c

Enter the coefficients for your equation:

√(2x + 3) = 5

Try these examples:

📊 Solution

Solution Found

x = 11

Step-by-Step Solution:

Original equation: √(2x + 3) = 5
Step 1: Square both sides: 2x + 3 = 5² = 25
Step 2: Subtract 3: 2x = 25 - 3 = 22
Step 3: Divide by 2: x = 22 / 2 = 11
Step 4: Verify - √(2(11) + 3) = √(25.0000) = 5.0000
✓ Solution verified: 5.0000 = 5

√ How to Solve Radical Equations

A radical equation contains a variable inside a radical symbol (like √ or ∛). Solving these equations requires isolating the radical and then eliminating it by raising both sides to an appropriate power.

Step-by-Step Method

  1. Isolate the radical on one side of the equation
  2. Raise both sides to the power of the index (square for √, cube for ∛)
  3. Solve the resulting equation
  4. Check all solutions in the original equation

Example: Solve √(2x + 3) = 5

√(2x + 3) = 5

2x + 3 = 25     ← square both sides

2x = 22     ← subtract 3

x = 11     ← divide by 2

Check: √(2(11) + 3) = √25 = 5 ✓

⚠️ Watch Out for Extraneous Solutions

Example of an extraneous solution:

√(x) = -3

x = 9     ← squaring gives this

But √9 = 3, not -3! So x = 9 is extraneous (no real solution exists).

📋 Key Rules

• (√a)² = a

• (∛a)³ = a

• √a exists only if a ≥ 0

• ∛a exists for all real a

• Always check solutions!

🔢 Perfect Squares

√1=1, √4=2, √9=3

√16=4, √25=5, √36=6

√49=7, √64=8, √81=9

√100=10, √121=11

Frequently Asked Questions

To solve radical equations: 1) Isolate the radical on one side of the equation. 2) Raise both sides to the power that eliminates the radical (square for square roots, cube for cube roots). 3) Solve the resulting equation for the variable. 4) ALWAYS check your answer by substituting back into the original equation to identify extraneous solutions.

Extraneous solutions are answers that emerge from the solving process but don't actually satisfy the original equation. They often appear when you square both sides of an equation because squaring can introduce false solutions. For example, if √x = -3, squaring gives x = 9, but √9 = 3 ≠ -3, so x = 9 is extraneous. Always verify your solutions!

√32 = √(16 × 2) = √16 × √2 = 4√2 ≈ 5.657. The simplified radical form is 4√2. For cube root: ∛32 = ∛(8 × 4) = 2∛4 ≈ 3.175.

Step 1: Square both sides: 2x + 3 = 25. Step 2: Subtract 3: 2x = 22. Step 3: Divide by 2: x = 11. Step 4: Check: √(2(11) + 3) = √25 = 5 ✓. The solution x = 11 is valid.

Yes! A radical equation has no solution when: 1) The equation requires a square root to equal a negative number (impossible for real numbers), like √x = -5. 2) All solutions found are extraneous after checking. For example, √(x-1) + x = 3 may have solutions that don't work when verified.

They are mathematically equivalent: √x = x^(1/2). Similarly, ∛x = x^(1/3), and the nth root of x equals x^(1/n). This relationship is useful because it allows you to apply exponent rules to radical expressions.

⚠️ Note: This calculator solves common types of radical equations with one variable. For more complex equations (like x + √x = 6), try isolating the radical and using quadratic methods, or consult a computer algebra system for step-by-step solutions.