Solving Radical Equations

Learning Objective

Solve algebraic equations with radical terms.

Introduction

When a radical expressionA quantity that contains a term with a radical, as in `2sqrt3` or `root(3)(8a^3bc^6)`. appears in an equation, we call that a radical equationAn equation that contains a variable within a radical term.. 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.

Squaring Both Sides

A basic strategy for solving radical equations is to isolate the radical term first, and then use the inverse operation (raising the radical term to a power) to pull out the variable. This is the same type of strategy we use to solve other, non-radical equations: rearrange the expression to isolate the variable we want to know, and then solve the resulting equation.

Let’s start with a radical equation that we can solve in a few steps: `sqrtx-3=5`.

Example
Problem	`sqrtx-3=5`
	`sqrtx-3+3=5+3`	Add `3` to both sides to isolate variable term
	`sqrtx=8`	Collect like terms
	`(sqrtx)^2=8^2`	Square both sides to remove variable from the radical
Answer	`x = 64`	Simplify

To check our solution, we can substitute `64` in for `x` in the original equation. Does `sqrt64-3=5`? Yes—the square root of `64` is `8`, and `8-3=5`.

Notice how we 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 more terms there are before squaring, the more additional terms will be generated by the process of squaring, and that gets messy fast.

Another warning: be careful not to just square individual terms in the equation. This method only works if we square both sides of the equation.

Even if we follow the rules, it’s not all sunshine and roses. Let’s try another problem that demonstrates a potential pitfall of squaring both sides to remove the radical:

Example
Problem	`sqrt(a-5)=-2`
	`(sqrt(a-5))^2=(-2)^2`	Square both sides to remove the term `a - 5` from the radical.
	`a-5 = 4`	Simplified equation
	`a-5 + 5 = 4 + 5`	Add `5` to both sides to isolate variable
Answer	`a = 9`	Collect like terms

We get a value of `9` for `a`. But watch what happens if we substitute `9` in for `a` in the original equation:

`sqrt(a-5)=-2`

`sqrt(9-5)=-2`

`sqrt4=-2`

This is incorrect—the square root of `4` is `2`, not `-2`. Our answer that `a = 9` does not check out. What happened?

Look back at the original problem: `sqrt(a-5)=-2`. Notice that the radical is set equal to `-2`, and recall that a square root of a number can only be positive. This means that no value for `a` will result in a radical expression whose square root is `-2`! We could have noticed that right away and concluded that there were no solutions for `a`. But why did the squaring method work in the first example and not in the second example?

The answer lies in the process of squaring itself. When we raise a number to an even power—whether second, fourth, or `50`^th power—we can introduce a false solution because the result of an even power is always a positive number. Think about it: `3^2` and `(-3)^2` are both `9`, and `2^4` and `(-2)^4` are both `16`. So when we squared `-2` and got `4` in this problem, we artificially turned the quantity positive. This is why we were still able to find a value for `a`—we solved the problem as if we were solving `sqrt(a-5)=2`! Oops.

Maritza is solving the following radical equation: `sqrth-4=2sqrth-7`.

Which of the following steps should Maritza do first?

A) Multiply both sides by `h`.

B) Square both sides.

C) Divide both sides by `sqrth-4`.

D) Subtract `sqrth` from both sides.

Show/Hide Answer

Roots Beyond Squares

The technique of eliminating a radical by raising both sides of an equation to the same power can be used with roots beyond squares as well. Consider the equation `-3=root(3)(b-2)`.

At first glance, this equation may seem unsolvable since the radical is set equal to `-3`. We don’t want to make that mistake again! But wait—this time the radical is not an even root—it is an odd root, `3`. This means that the radical expression can be set equal to a negative number since radicals with odd roots may have negative values as roots.

Example
Problem	`-3=root(3)(b-2)`
	`(-3)^3=(root(3)(b-2))^3`	Cube both sides to remove the radical.
	`-27 = b - 2`	Simplified equation. Note that cubing preserves the negative sign in front of `27`.
	`-27 + 2 = b - 2 + 2`	Add `2` to both sides to isolate variable.
Answer	`-25 = b`	Collect like terms.

As with all answers we get when we solve radical equations, we should substitute `-25` in for `b` in our original equation to make sure it is a legitimate solution.

`-3=root(3)(b-2)`

`-3=root(3)(-25-2)`

`-3=root(3)(-27)`

`-3 = -3`

The substitution results in a true statement, so our answer is right.

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 negative terms will be artificially changed into positive terms. When solving these problems, it is important to check the answer by substituting the value back into the original equation.