What if there are two radicals in the equation?

Isolate one radical on one side, then raise both sides to the appropriate power. This will eliminate one radical but may leave another. Isolate the remaining radical and repeat the process.

Why do extraneous solutions occur?

Raising both sides of an equation to an even power can introduce new solutions because it eliminates sign information. For instance, both $x = 2$ and $x = -2$ yield $x^2 = 4$ when squared.

Can a radical equation have no solutions?

Yes. If all calculated values are extraneous, the equation has no real solution. For example, $\sqrt{x} = -5$ has no real solution because the principal square root is defined as non-negative.

How to solve a radical equation

To solve a radical equation, isolate the radical expression on one side of the equal sign and raise both sides to the power of the index. This method applies whenever the variable you need to solve for is contained within a square root, cube root, or higher-order root.

The setup

Identify the radical containing the variable. Move all other terms to the opposite side of the equation using standard algebraic operations.

The steps

Isolate the radical on one side of the equation. 2. Raise both sides of the equation to the power of the index. For example, square both sides for a square root. 3. Solve the resulting polynomial equation. 4. If there are multiple radicals, you may need to repeat steps 1 and 2.

Checking the result

Substitute each potential solution back into the original equation. Solutions that result in a false statement are extraneous and must be discarded. In the real number system, the principal even root of a number cannot be negative.

Common errors

Failing to check for extraneous solutions is the most frequent error. Another common mistake is squaring a binomial incorrectly, such as assuming $(a+b)^2 = a^2 + b^2$ instead of using the correct expansion $(a+b)^2 = a^2 + 2ab + b^2$ .

Worked example

Solve the equation $\sqrt{2x + 7} - x = 2$ .

First, isolate the radical: $\sqrt{2x + 7} = x + 2$ . Next, square both sides: $(\sqrt{2x + 7})^2 = (x + 2)^2$ . Expand the right side: $2x + 7 = x^2 + 4x + 4$ . Move all terms to one side to form a quadratic equation: $x^2 + 2x - 3 = 0$ . Factor the quadratic: $(x + 3)(x - 1) = 0$ . The potential solutions are $x = -3$ and $x = 1$ . Check $x = -3$ in the original equation: $\sqrt{2(-3) + 7} - (-3) = \sqrt{1} + 3 = 1 + 3 = 4 eq 2$ . Thus, $-3$ is extraneous. Check $x = 1$ in the original equation: $\sqrt{2(1) + 7} - 1 = \sqrt{9} - 1 = 3 - 1 = 2$ . This is correct. The only valid solution is $x = 1$ .

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References: OpenStax College Algebra, Chapter 2: Equations and Inequalities · Khan Academy: Solving radical equations