Square Root of An Equation Calculator
'/%3E%3Ccircle cx='48' cy='39' r='19' fill='%23f2c9a5'/%3E%3Cpath d='M27 35c3-16 39-17 42 2-8-8-27-9-42-2z' fill='%23374151'/%3E%3Cpath d='M21 86c5-24 49-28 56 0z' fill='%232563eb'/%3E%3Ccircle cx='41' cy='41' r='2' fill='%23111827'/%3E%3Ccircle cx='55' cy='41' r='2' fill='%23111827'/%3E%3Cpath d='M42 52c4 3 9 3 13 0' stroke='%2392400e' stroke-width='3' fill='none' stroke-linecap='round'/%3E%3C/svg%3E)
Reviewed by Calculator Editorial Team
This calculator helps you find the square root of an equation. Whether you're solving quadratic equations or working with radicals, understanding how to find square roots is essential in algebra and calculus.
What is the Square Root of an Equation?
The square root of an equation refers to finding the value(s) of x that satisfy the equation when the variable is under a square root. For example, in the equation √(x + 5) = 3, we need to find x such that the square root of (x + 5) equals 3.
Key Formula: If √(f(x)) = k, then f(x) = k².
Solving such equations involves squaring both sides to eliminate the square root, then solving for the variable. It's important to check solutions for extraneous roots that may result from squaring both sides.
How to Solve Square Roots of Equations
Step 1: Isolate the Square Root
Move all terms not involving the square root to the other side of the equation. For example, in √(2x + 1) - 5 = 0, add 5 to both sides to get √(2x + 1) = 5.
Step 2: Square Both Sides
Eliminate the square root by squaring both sides of the equation. In our example, squaring both sides gives 2x + 1 = 25.
Step 3: Solve for the Variable
Isolate the variable by performing inverse operations. In our example, subtract 1 and divide by 2 to get x = 12.
Step 4: Check for Extraneous Solutions
Always verify solutions by plugging them back into the original equation. In our example, √(2*12 + 1) = √25 = 5, which matches the right side.
Note: Squaring both sides can introduce extraneous solutions that don't satisfy the original equation. Always check your solutions.
Worked Examples
Example 1: Simple Square Root Equation
Solve √(3x - 2) = 4.
- Square both sides: 3x - 2 = 16
- Add 2: 3x = 18
- Divide by 3: x = 6
- Check: √(3*6 - 2) = √16 = 4 ✓
Example 2: Equation with Fractional Coefficient
Solve √(5x + 3) = 2.
- Square both sides: 5x + 3 = 4
- Subtract 3: 5x = 1
- Divide by 5: x = 0.2
- Check: √(5*0.2 + 3) = √4 = 2 ✓
| Equation | Solution | Verification |
|---|---|---|
| √(3x - 2) = 4 | x = 6 | √16 = 4 ✓ |
| √(5x + 3) = 2 | x = 0.2 | √4 = 2 ✓ |
FAQ
- What is the difference between solving √x = a and x = a²?
- The equation √x = a has solutions x = a², but only if a ≥ 0. The equation x = a² has solutions x = a, but only if x ≥ 0. The square root function is defined only for non-negative numbers.
- Can I solve equations with √(x) + √(y) = k?
- Yes, but it's more complex. You would need to isolate one square root, square both sides, then isolate the other square root and square again. This often leads to extraneous solutions that need verification.
- What happens if the equation has no real solutions?
- If the expression inside the square root is negative, there are no real solutions. For example, √(x - 5) = 3 has no real solutions when x < 5 because the square root of a negative number isn't real.
- How do I solve equations with nested square roots like √(√x + 2) = 3?
- First, isolate the outer square root: √x + 2 = 9. Then subtract 2: √x = 7. Finally, square both sides: x = 49. Always check solutions to ensure they satisfy the original equation.