Radical math solver: solve square root equations

How a radical equation is solved

The method has four moves. First get the square root alone on one side. Second, square both sides, which removes the root. Third, solve the equation that results, often a quadratic. Fourth, put every answer back into the original equation. This radical math solver follows those moves and writes the check as its own step.

Why the check is not optional

Squaring is not reversible. The equations x = 3 and x = −3 are different, but both become x² = 9 when squared. So after squaring, the new equation can have answers the original does not. Two things are always true about the original: the expression under the root must be zero or more, and the value of a square root is never negative. In sqrt(x + 6) = x the right side is the root, so x must be 0 or more. The squared equation gives x = 3 and x = −2, and the second one fails: sqrt(4) is 2, not −2. Only x = 3 is a solution.

How to use it

1. Type the square root as sqrt( ) with the whole expression inside the brackets, for example sqrt(x + 5) + 1 = x.
2. Read how the solver isolates the root before squaring.
3. Look at the check for extra answers. Each answer is put back into the original equation.
4. Keep only the answers that pass, and make sure the expression under the root is not negative.

Radical math solver: worked examples with checks for extra answers

Each example is the solver's own output for the equation shown. We also worked every one by hand, and the answers match. Use the link under an example to run it on the board above.

1. Solve sqrt(3x + 1) = 5

1. Square both sides(sqrt(3x + 1))^2 = 5^2Squaring both sides eliminates the square root on the left. The root must be defined, so 3x + 1 must be at least 0 (x ≥ −1/3); the right side, 5, is already non-negative.
2. Simplify both sides3x + 1 = 25Evaluate the square on both sides.
3. Subtract 1 from both sides3x = 24Subtract 1 from both sides to isolate the term with the variable.
4. Divide by 3x = 8Divide both sides by 3 to solve for x.

Answer: x = 8

Check: Substitute x = 8 back into the original equation: sqrt(3 * 8 + 1) = sqrt(25) = 5, which matches the right-hand side.

Open this example on the board above →

2. Solve sqrt(x + 6) = x

1. Square both sides to eliminate the square root(sqrt(x + 6))^2 = x^2 implies x + 6 = x^2Squaring both sides removes the radical. The root needs x + 6 ≥ 0 and its value is never negative, so any answer must also satisfy x ≥ 0. Squaring can add solutions that break this, so we check at the end.
2. Rearrange terms to set the quadratic equation to zerox^2 - x - 6 = 0Subtract x and 6 from both sides to write the quadratic equation in standard form.
3. Factor the quadratic expression(x - 3)(x + 2) = 0Find two numbers that multiply to -6 and add to -1, which are -3 and 2.
4. Solve for possible values of xx = 3 or x = -2Set each linear factor equal to zero using the zero product property.
5. Check for extraneous solutionsFor x = -2: sqrt(-2 + 6) = sqrt(4) = 2 != -2; For x = 3: sqrt(3 + 6) = sqrt(9) = 3 = 3The principal square root must be non-negative, so x = -2 is an extraneous solution, leaving only x = 3.

Answer: x = 3

Check: Substituting x = 3 into the original equation yields sqrt(3 + 6) = sqrt(9) = 3, which equals the right-hand side.

Open this example on the board above →

3. Solve sqrt(x + 5) + 1 = x

1. Isolate the radical termsqrt(x + 5) = x - 1Subtract 1 from both sides to isolate the square root expression on one side. A square root is never negative, so any answer must satisfy x − 1 ≥ 0, that is x ≥ 1 (and x + 5 ≥ 0).
2. Square both sides to eliminate the radicalx + 5 = (x - 1)^2Squaring both sides removes the radical, keeping in mind that we must check for extraneous solutions later.
3. Expand the right-hand sidex + 5 = x^2 - 2x + 1Expand the squared binomial using the identity (a - b)^2 = a^2 - 2ab + b^2.
4. Rearrange into standard quadratic formx^2 - 3x - 4 = 0Subtract x and 5 from both sides to set the quadratic equation equal to zero.
5. Factor the quadratic equation(x - 4)(x + 1) = 0Find two numbers that multiply to -4 and add to -3, which are -4 and 1.
6. Solve for potential values of xx = 4 or x = -1Apply the zero-product property by setting each factor equal to zero.
7. Check for extraneous solutionsFor x = -1: sqrt(-1 + 5) + 1 = 3 != -1; For x = 4: sqrt(4 + 5) + 1 = 4Substitute both values back into the original equation. x = -1 is extraneous, leaving x = 4 as the only valid solution.

Answer: x = 4

Check: Substitute x = 4 into the original equation: sqrt(4 + 5) + 1 = sqrt(9) + 1 = 3 + 1 = 4, which matches the right-hand side.

Open this example on the board above →

Mistakes the steps help you catch

• Squaring before the root is alone, which leaves a root behind.
• Squaring the two terms of a sum separately: (1 + sqrt(x))² is 1 + 2 sqrt(x) + x, not 1 + x.

If the squared equation is a quadratic, the factoring solver shows how to split it. For other equations, see solve for x.

FAQ

Related solvers