An equation that contains a radical expression is called a radical equation. Solving radical equations requires applying the rules of exponents and following some basic algebraic principles. In some cases, it also requires looking out for errors generated by raising unknown quantities to an even power.

Solve an equation containing a single square root

A basic strategy for solving radical equations is to isolate the radical term first, and then raise both sides of the equation to a power to remove the radical.  Let’s look at two properties we will use in this process.  First, if [latex] a=b[/latex], then [latex] {{a}^{2}}={{b}^{2}}[/latex]. This property allows you to square both sides of an equation and remain certain that the two sides are still equal.  Second, if the square root of any nonnegative number x is squared, then you get x: [latex] {{\left( \sqrt{x} \right)}^{2}}=x[/latex]. This property allows you to “remove” the radicals from your equations.

Let’s start with a radical equation that you can solve in a few steps:[latex] \sqrt{x}-3=5[/latex].

To check your solution, you can substitute [latex]64[/latex] in for [latex]x[/latex] in the original equation. Does [latex] 2\sqrt{64}-6=10[/latex]? Yes—the principle square root of [latex]64[/latex] is [latex]8[/latex], and [latex]2(8)−6=10[/latex].

Notice how you combined like terms and then squared both sides of the equation in this problem. This is a standard method for removing a radical from an equation. It is important to isolate a radical on one side of the equation and simplify as much as possible before squaring. The fewer terms there are before squaring, the fewer additional terms will be generated by the process of squaring.

In the example above, only the variable x was underneath the radical. Sometimes you will need to solve an equation that contains multiple terms underneath a radical. Follow the same steps to solve these, but pay attention to a critical point—square both sides of an equation, not individual terms. Watch how the next two problems are solved.

In the following video we show how to solve simple radical equations.

Identify extraneous solutions

Following rules is important, but so is paying attention to the math in front of you—especially when solving radical equations. Take a look at this next problem that demonstrates a potential pitfall of squaring both sides to remove the radical.

Look at that—the answer [latex]a=9[/latex] does not produce a true statement when substituted back into the original equation. What happened?

Check the original problem: [latex]\sqrt{a-5}=-2[/latex]. Notice that the radical is set equal to [latex]−2[/latex], and recall that the principal square root of a number can only be positive. This means that no value for a will result in a radical expression whose positive square root is [latex]−2[/latex]! You might have noticed that right away and concluded that there were no solutions for a.

Incorrect values of the variable, such as those that are introduced as a result of the squaring process are called extraneous solutions. Extraneous solutions may look like the real solution, but you can identify them because they will not create a true statement when substituted back into the original equation. This is one of the reasons why checking your work is so important—if you do not check your answers by substituting them back into the original equation, you may be introducing extraneous solutions into the problem.
In the next video example, we solve more radical equations that may have extraneous solutions.

Summary

A common method for solving radical equations is to raise both sides of an equation to whatever power will eliminate the radical sign from the equation. But be careful—when both sides of an equation are raised to an even power, the possibility exists that extraneous solutions will be introduced. When solving a radical equation, it is important to always check your answer by substituting the value back into the original equation. If you get a true statement, then that value is a solution; if you get a false statement, then that value is not a solution.