Radical Equation Solver

Results

Solution

x = 9

for √(ax + b) = c

Equation	√(1·x + 0) = 3
Square both sides	1·x + 0 = 9
Solve for x	x = (c² − b) / a = 9

What is a radical equation?

A radical equation contains a variable inside a square root. This solver handles equations of the form $\sqrt{ax + b} = c$, where a, b, and c are real numbers. It isolates the radical, squares both sides, and solves the resulting linear equation for x, while checking whether a real solution actually exists.

Number line showing the principal square root output is always zero or positive — Because √ returns a non-negative value, the equation has no solution when c is negative.

How to use it

Enter the coefficient a (multiplying x inside the root), the constant b (added inside the root), and c (the value on the right of the equals sign). Press calculate to get x along with a step-by-step breakdown. If c is negative or a is zero, the tool reports that no real solution exists.

The formula explained

Starting from $\sqrt{ax + b} = c$, square both sides to remove the radical: $ax + b = c^{2}$. Subtract b and divide by a to get $$x = \frac{c^{2} - b}{a}$$ Because the square root function only returns non-negative values, the original equation can only be true when $c \geq 0$. If c is negative there is no real x that satisfies it. The coefficient a must also be non-zero, otherwise x disappears from the equation.

Diagram showing the steps to solve square root of a x plus b equals c by squaring both sides — Squaring both sides turns √(ax+b)=c into ax+b=c², giving x=(c²−b)/a.

Worked example

Solve $\sqrt{2x + 1} = 3$. Here $a = 2$, $b = 1$, $c = 3$. Square both sides: $2x + 1 = 9$. Then $2x = 8$, so $x = 4$. Check: $\sqrt{2 \cdot 4 + 1} = \sqrt{9} = 3$. ✓

FAQ

Why does a negative c give no solution? The principal square root is always $\geq 0$, so it can never equal a negative number.

What if a = 0? Then the equation has no x term to solve, so the tool returns no solution.

Should I always check my answer? Yes. Squaring can introduce extraneous roots, so substitute x back into the original equation to confirm.

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