An equation that contains a radical expression is called a radical equation. A simple example of a radical equation is

[latex]\sqrt{x}=5[/latex]

As with many other kinds of equations, since the goal is to solve for [latex]x,[/latex] we must remove the radical sign. We have seen earlier in this chapter that [latex](\sqrt{x})^2=x[/latex], so squaring both sides will solve the equation:

[latex](\sqrt{x})^2=5^2[/latex]

The solution is [latex]25[/latex]. We can check our solution and see that the principal square root of [latex]25[/latex] is indeed [latex]5.[/latex]

We should clarify that raising both sides of an equation to a power is indeed a valid equation-solving step:

We can now use this basic strategy in more complicated equations.

Isolate a Radical Term

The key to solving more complicated 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 us start with a radical equation that you can solve in a few steps.

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 [latex]x[/latex] 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 example, you will see two more examples that are similar to the ones above.

Remember that if an equation to solve has rational exponents in it, these are equivalent to radicals so they can be solved the same way. You can either change them to radicals first or just leave them as exponents, as we do here.

Identify a Radical Equation with No Solutions or 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 nonnegative. This means that no value for [latex]a[/latex] will result in a radical expression whose principal square root is [latex]−2[/latex]! You might have noticed that right away and concluded that there were no solutions for [latex]a[/latex]. But why did the process of squaring create an answer, [latex]a=9[/latex], that proved to be incorrect?

The answer lies in the process of squaring itself. When you raise a number to an even power—whether it is the second, fourth, or [latex]50[/latex]th power—you can introduce a false solution because the result of an even power is always a nonnegative number. Think about it: if you begin with the false statement [latex]-3=3[/latex] and square both sides, you get a true statement [latex]9=9[/latex], so squaring both sides can turn a false statement true. When you squared [latex]−2[/latex] and got [latex]4[/latex] in the previous problem, you artificially turned the quantity positive. This is why you were still able to find a value for [latex]a[/latex]—you solved the problem as if you were solving[latex] \sqrt{a-5}=2[/latex]! (The correct solution to [latex] \sqrt{a-5}=-2[/latex] is actually “no solution.”)

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 following video, we present more examples of solving radical equations by isolating a radical term on one side.

Have a look at the following problem. Since the left side has a variable in it, we have no way of knowing before starting the problem whether any solutions will be extraneous or not. Our only option is to check each solution at the end in the original equation.

It may be difficult to understand why extraneous solutions exist at all. Thinking about extraneous solutions by graphing the equation may help you make sense of what is going on.

You can graph [latex]x+4=\sqrt{x+10}[/latex] on a coordinate plane by breaking it into a system of two equations: [latex]y=x+4[/latex] and [latex]y=\sqrt{x+10}[/latex]. The graph is shown below. Notice how the two graphs intersect at one point, when the value of [latex]x[/latex] is [latex]−1[/latex]. This is the value of [latex]x[/latex] that satisfies both equations, so it is the solution to the system.

Now, following the work we did in the example problem, let us square both of the expressions to remove the variable from the radical. Instead of solving the equation [latex]x+4=\sqrt{x+10}[/latex], we are now solving the equation [latex]\left(x+4\right)^{2}=\left(\sqrt{x+10}\right)^{2}[/latex] , or [latex]x^{2}+8x+16=x+10[/latex]. The graphs of [latex]y=x^{2}+8x+16[/latex] and [latex]y=x+10[/latex] are plotted below. Notice how the two graphs intersect at two points, when the values of [latex]x[/latex] are [latex]−1[/latex] and [latex]−6[/latex].

Although [latex]x=−1[/latex] is shown as a solution in both graphs, squaring both sides of the equation had the effect of adding an extraneous solution, [latex]x=−6[/latex]. Again, this is why it is so important to check your answers when solving radical equations!

In addition to both solutions checking like in the previous example, sometimes you get two solutions and neither of them checks. In this case we would again say the equation has no solution.

In the next video, we show more examples of radical equations that have extraneous solutions. Always remember to check your answers for radical equations.

Two Radical Terms

In our last examples, we have two radical terms in the starting equation. The process is similar, but since we have two radicals we can typically only isolate one of them at a time. Then squaring both sides will remove that radical.

In this video, we show the solution for another radical equation that has square roots on both sides of the equation. Pay special attention to how you square the side that has the sum of a radical term and a constant.

Sometimes each side of the equation has only a single radical expression. These examples are simpler because squaring eliminates both radicals simultaneously.

Applications

Kinetic Energy

In physics, the kinetic energy of an object represents the amount of energy it has due to motion, and how much energy it would take to stop the object.

Here is another example of finding the kinetic energy of an object in motion.

Volume

Harvester ants found in the southwest of the U.S. create a vast interlocking network of tunnels for their nests. As a result of all this excavation, a very common above-ground hallmark of a harvester ant nest is a conical mound of small gravel or sand ^[1]

Here is another example of finding volume given the radius of a cone.

Summary

To solve a radical equation, first isolate the radical and then raise both sides of the equation to whatever power will eliminate the radical symbol 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.

Radical equations play a significant role in science, engineering, and even music. Sometimes you may need to use what you know about radical equations to solve for different variables in these types of problems.