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Radical Equation Calculator

Solve √(ax + b) = c for x by squaring both sides. The calculator shows each step and checks whether the solution is valid or extraneous.

Solve: √(a·x + b) = c

Result

x = 11

sqrt(2x + 3) = 5 gives x = 11, a valid solution

Quick Answer

A radical equation calculator solves √(ax + b) = c by squaring both sides to get ax + b = c², then solving for x = (c² − b) / a. For √(3x + 7) = 4: c² = 16, so x = (16 − 7)/3 = 9/3 = 3. Check: √(3 × 3 + 7) = √16 = 4. Valid solution: x = 3.

Radical Equations: What Squaring Both Sides Does and Why a Check Is Required

This calculator solves equations of one specific form: √(ax + b) = c, where the entire expression under the radical is linear. Squaring both sides removes the square root and converts the radical equation into a simple linear equation. However, squaring is not a reversible operation. It can introduce extraneous solutions: values of x that satisfy the squared equation but do not satisfy the original radical equation.

The Radical Equation Formula: Step-by-Step Derivation

√(ax + b) = c → ax + b = c² → x = (c² − b) / a Condition: c ≥ 0 and a ≠ 0

  • √(ax + b) = c
  • Condition 1: c ≥ 0 (square root cannot equal a negative number)
  • Squared: ax + b = c²
  • Solution: x = (c² − b) / a
  • Condition 2: substitute x back into original equation to verify not extraneous

Solve a Radical Equation in 4 Steps

Inputs

  • a: the coefficient of x under the square root sign
  • b: the constant term under the square root sign
  • c: the right-hand side value (must be ≥ 0)

Steps

  1. Enter a, b, and c.
  2. The calculator checks that c ≥ 0 and a ≠ 0.
  3. It computes x = (c² − b) / a.
  4. It substitutes x into √(ax + b) and compares to c; if the check passes, x is the valid solution. If not, it reports an extraneous solution.

Radical Equation Worked Example: √(2x − 5) = 3

Solving the radical equation √(2x − 5) = 3 step by step.

  1. Check c ≥ 0: c = 3 ≥ 0 (valid).
  2. Square both sides: 2x − 5 = 3² = 9.
  3. Solve for x: 2x = 9 + 5 = 14 → x = 7.
  4. Extraneous check: √(2 × 7 − 5) = √9 = 3. Matches c = 3.

x = 7 is the valid, non-extraneous solution.

When This Radical Equation Calculator Applies

This calculator solves only equations of the form √(ax + b) = c, a single square root on the left and a constant on the right. It applies to introductory radical equation problems in algebra II and pre-calculus.

Assumptions

  • The equation has exactly the form √(ax + b) = c.
  • a ≠ 0.
  • c ≥ 0; if c < 0, there is no real solution.
  • The square root is the principal (non-negative) square root.

Limitations

  • Does not solve equations with radicals on the right side, both sides, or with nested radicals.
  • Does not handle higher-order roots (cube root, fourth root, etc.).
  • Does not solve radical equations with quadratic expressions under the radical.

In Practice

An extraneous solution arises most often when b is negative and c is small. For example, √(x − 10) = −2 has no real solution (c < 0), but if you ignored the c < 0 check and squared both sides, you would get x − 10 = 4, x = 14, and then verify √(14 − 10) = √4 = 2 ≠ −2. The check reveals the extraneous result. Always perform the substitution check.

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Frequently Asked Questions: Radical Equation Calculator

Why do radical equations produce extraneous solutions?

Squaring both sides of an equation is not an equivalence operation: it can introduce solutions that do not satisfy the original. Specifically, squaring eliminates sign information: the equation (√(ax+b))² = c² is satisfied by both √(ax+b) = c and √(ax+b) = −c, even though only the first is true for a principal square root. The extraneous solution is the one introduced by the negative case.

What if c is negative?

If c < 0, there is no real solution. The principal square root of any real number is always non-negative; it cannot equal a negative number. The equation √(ax + b) = c with c < 0 is impossible in the real numbers.

What does extraneous solution mean?

An extraneous solution is a value of x that satisfies the squared equation but not the original radical equation. It appears in the algebra but is not a true solution. The substitution check reveals it.

What if a = 0 in √(ax + b) = c?

If a = 0, the equation becomes √b = c, which has nothing to do with x. It is either true for all x (if √b = c) or false for all x (if √b ≠ c). The calculator treats a = 0 as an undefined input.

Can I solve equations with two radicals using this calculator?

No. Equations such as √(ax + b) + √(cx + d) = e require isolating one radical, squaring, isolating the remaining radical, and squaring again: two squaring steps that introduce additional extraneous-solution risk.

Sources

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