Basic Radical Equations

A radical equation is an equation where the variable is located inside a radical, such as a square root. To solve these equations, the main goal is to eliminate the radical so you can solve for the variable.

The 4-Step Process

Solving a basic radical equation involves four key steps:

1. Isolate the radical: Make sure the radical expression is by itself on one side of the equal sign.
2. Square both sides: Raise both sides of the equation to the power of the index (for square roots, you square both sides) to eliminate the radical.
3. Solve the new equation: Solve the resulting linear or quadratic equation.
4. Check your answers: This step is mandatory! Squaring both sides of an equation can introduce extraneous solutions—mathematical answers that do not actually work in the original equation.

Example 1: A Simple Radical Equation

Solve: $\sqrt{x + 3} = 5$

Step 1: Isolate the radical. The radical $\sqrt{x + 3}$ is already isolated on the left side.

Step 2: Square both sides. $(\sqrt{x + 3})^2 = (5)^2$ $x + 3 = 25$

Step 3: Solve the new equation. Subtract 3 from both sides: $x = 22$

Step 4: Check the answer. Plug $x = 22$ back into the original equation: $\sqrt{22 + 3} = 5$ $\sqrt{25} = 5$ $5 = 5$ The solution is valid. Final answer: $x = 22$.

Example 2: Extraneous Solutions

Solve: $\sqrt{2x - 1} = x - 2$

Step 1: Isolate the radical. The radical is already isolated.

Step 2: Square both sides. $(\sqrt{2x - 1})^2 = (x - 2)^2$ $2x - 1 = x^2 - 4x + 4$

Step 3: Solve the new equation. Since we have an $x^2$ term, this is a quadratic equation. Move all terms to one side to set the equation to zero. Subtract $2x$ and add $1$ to both sides: $0 = x^2 - 6x + 5$

Factor the quadratic equation: $0 = (x - 1)(x - 5)$

Our potential solutions are $x = 1$ and $x = 5$.

Step 4: Check for extraneous solutions. Let's test both potential solutions in the original equation.

Check $x = 1$: $\sqrt{2(1) - 1} = 1 - 2$ $\sqrt{1} = -1$ $1 = -1$ (False!) Since the principal square root of 1 is positive 1, this does not work. $x = 1$ is an extraneous solution.

Check $x = 5$: $\sqrt{2(5) - 1} = 5 - 2$ $\sqrt{9} = 3$ $3 = 3$ (True!)

The only valid solution is $x = 5$.