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Radical equations

Radical Equation Calculator

Enter an equation with one square root, typed as sqrt(x + 3) = x - 3 or √(x + 3) = x - 3. The calculator squares both sides, solves, and checks each answer.

Preview sqrt(x + 3) = x - 3 (Solve) Ready
Examples
Solution

Worked example: sqrt(x + 3) = x - 3

Worked example. Edit the problem above and press Solve to solve your own.

Problemsqrt(x + 3) = x - 3
  1. Isolate the square root

    Get the square root by itself on one side before squaring.

    sqrt(x + 3) = x - 3
    Original equation
    √(x + 3) = x − 3
    Square root alone

    Adding, subtracting, multiplying, or dividing both sides by the same nonzero number keeps the equation balanced.

  2. Square both sides

    Squaring removes the square root. It can also add answers that do not work, so every answer gets checked at the end.

    x + 3 = (x − 3)²
    Square both sides
    x + 3 = x² − 6x + 9
    Multiply out
    x² − 7x + 6 = 0
    Everything on one side

    If two quantities are equal, their squares are equal too.

  3. Write the quadratic in standard form

    Move every term to the left and combine like terms, then identify a, b, and c.

    x² − 7x + 6 = 0
    Original equation
    x² − 7x + 6 = 0
    Standard form: ax² + bx + c = 0
    a = 1, b = −7, c = 6
    Identify the coefficients

    Standard form exposes the exact coefficients used by complete quadratic methods.

  4. Compute the discriminant

    Substitute the exact coefficients into D = b² − 4ac.

    D = b² − 4ac
    Discriminant formula
    D = (−7)² − 4(1)(6)
    Substitute a, b, and c
    D = 25
    Simplify

    The discriminant determines the number and type of roots.

  5. Apply the quadratic formula and check the roots

    Substitute the exact coefficients, simplify both roots, and check rational candidates.

    x = (−b ± √(b² − 4ac)) / 2a
    Quadratic formula
    x = (7 ± √25) / 2
    Substitute the exact values
    x = 1 or x = 6
    Simplify the root values
    x = 6: Exact residual: 0
    Substitution verified
    x = 1: Exact residual: 0
    Substitution verified

    For a nonzero quadratic coefficient, the formula enumerates the complete root set.

  6. Check each answer in the original equation

    Substitute every candidate into the original equation. Keep it only if the square root equals the other side.

    x = 1 ✗ extraneous
    √(4) and −2 differ: the right side is negative
    x = 6 ✓
    √(9) and 3 are equal

    Squaring can create extraneous solutions, so this check decides the final answer.

Answerx = 6

Radical equations guide

How the radical equation calculator solves square root equations

This radical equation calculator solves equations with one square root, like √(x + 3) = x − 3. It isolates the root, squares both sides, solves, and checks every answer for extraneous solutions.

How do you solve a radical equation?

Get the square root alone on one side, then square both sides to remove it. Solve the equation that results, which is often a quadratic. Then substitute every answer into the original equation and reject any that fail. For √(x + 3) = x − 3, squaring gives x = 1 or x = 6, but only x = 6 checks.

What is a radical equation?

A radical equation is an equation with the variable inside a root, such as √(2x + 1) = x − 1. An equation like x = √5 doesn't count, since the root contains only a number.

To enter one, type sqrt( ) around the expression under the root, or paste the √ symbol with brackets. Both sqrt(x + 3) = x - 3 and √(x + 3) = x - 3 work.

What are the steps for solving radical equations?

Isolate, square, solve, check. The last step is required, not optional, because squaring can add answers that don't work.

  1. Move everything except the square root to the other side.
  2. Square both sides. The root disappears on the left.
  3. Multiply out the right side and collect everything on one side.
  4. Solve the result, usually by factoring or the quadratic formula.
  5. Substitute each answer into the original equation and keep only the ones that work.
Start
√(x + 3) = x − 3
Square both sides
x + 3 = (x − 3)²
Multiply out
x + 3 = x² − 6x + 9
Standard form
x² − 7x + 6 = 0
Solve
x = 1 or x = 6
Check x = 1
√4 = 2, but 1 − 3 = −2, so reject
Check x = 6
√9 = 3 and 6 − 3 = 3, so keep

The final answer is x = 6. The value x = 1 is extraneous: it solves the squared equation but not the original one.

Why does squaring create extraneous solutions?

Squaring hides the sign of each side. Two numbers that are different can have the same square, so the squared equation accepts answers the original rejects.

The simplest case is −2 = 2. That's false, but squaring both sides gives 4 = 4, which is true. The same thing happened above with x = 1: the left side was 2, the right side was −2, and their squares matched.

A principal square root, written √, is never negative. So any candidate that makes the other side negative is extraneous. That's a quick way to spot them before you finish the arithmetic.

Why do you isolate the square root before squaring?

If another term sits next to the root, squaring doesn't remove the root. Instead you get a messier equation with the root still inside a middle term.

For √(x + 7) + 5 = x, subtract 5 first to get √(x + 7) = x − 5. Squaring gives x + 7 = x² − 10x + 25, so x² − 11x + 18 = 0 and x = 2 or x = 9. The check rejects x = 2, since √9 = 3 while 2 − 5 = −3. The answer is x = 9.

A number multiplied by the root is fine to leave in place. For 2√(x + 1) = x − 2, squaring both sides gives 4(x + 1) = x² − 4x + 4, so x² − 8x = 0 and x = 0 or x = 8. At x = 0 the left side is 2 and the right side is −2, so the solver keeps only x = 8. The solver shows exactly these steps.

How do you square the other side correctly?

Treat it as a binomial times itself: (a − b)² = a² − 2ab + b². The middle term is the one people forget.

Side to squareSquared correctlyCommon mistake
x − 3x² − 6x + 9x² + 9 or x² − 9
x − 1x² − 2x + 1x² + 1
x − 5x² − 10x + 25x² + 25
2√(x + 1)4(x + 1)2(x + 1)

The last row is the other trap. When a number multiplies the root, square the number too. For √(2x + 1) = x − 1, squaring correctly gives 2x + 1 = x² − 2x + 1, so x² − 4x = 0. That gives x = 0 or x = 4, and the check leaves x = 4.

When does a radical equation have no solution?

It has no solution when every candidate fails the check, or when the isolated root equals a negative number. A principal square root can't be negative, so √(something) = −2 is impossible.

Take √x + 2 = 0. Subtracting 2 gives √x = −2, and the calculator stops right there with No solution. Squaring anyway would give x = 4, which fails the check since √4 = 2 and 2 + 2 is 4, not 0.

What kinds of radical equations can this solver handle?

The radical equation solver handles one square root in an equation, with a linear or quadratic expression on the other side. After squaring, it solves the result the same way the quadratic equation solver does, so answers like x = (−1 + √17)/2 for √(x + 5) − 1 = x come out exact.

You can use any letter for the variable, so √(t + 3) = t − 3 works the same way. Every answer comes with the squared equation, the candidates it produced, and a check line for each candidate that says whether it was kept or rejected and why.

Two things are outside its range. Equations with two or more square roots, such as √x + √(x + 5) = 5, aren't supported yet. Cube roots inside an equation, like ∛(x + 1) = 2, aren't either. For simplifying a single radical such as √72, use the simplify radicals calculator instead.

Frequently asked questions

Short answers to the questions people ask most about this topic.

What does a radical equation calculator do?

It solves an equation with x under a square root. It isolates the root, squares both sides, solves the result, and checks each answer in the original equation so extraneous solutions are removed.

Is a square root equation solver the same thing?

Yes. A square root equation is the most common kind of radical equation, and that's the kind this page solves. Type it as sqrt(x + 3) = x - 3.

What is an extraneous solution?

It's an answer that solves the squared equation but fails in the original. For √(x + 3) = x − 3, x = 1 is extraneous because the left side is 2 and the right side is −2.

Do I always have to check my answers?

Yes. Squaring both sides can add answers, so every candidate needs a check. Only the original equation decides which ones are real solutions.

Why does √x = −2 have no solution?

The √ symbol means the principal square root, which is never negative. No real x makes it equal −2.

Can it solve equations with two square roots?

Not yet. Equations like √x + √(x + 5) = 5 need two rounds of squaring, and this solver handles one square root per equation.

How is solving radical equations different from simplifying radicals?

Solving finds the x that makes an equation true. Simplifying rewrites a single root in a neater form, like √72 = 6√2. The simplify radicals calculator does the second job.

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