8.6 Solve Radical Equations

Solve Radical Equations

In this section we will solve equations that have a variable in the radicand of a radical expression. An equation of this type is called a radical equation.

As usual, when solving these equations, what we do to one side of an equation we must do to the other side as well. Once we isolate the radical, our strategy will be to raise both sides of the equation to the power of the index. This will eliminate the radical.

Solving radical equations containing an even index by raising both sides to the power of the index may introduce an algebraic solution that would not be a solution to the original radical equation. Again, we call this an extraneous solution as we did when we solved rational equations.

In the next example, we will see how to solve a radical equation. Our strategy is based on raising a radical with index n to the n^th power. This will eliminate the radical.

$For a \geq 0, {(\sqrt[n]{a})}^{n} = a .$

When we use a radical sign, it indicates the principal or positive root. If an equation has a radical with an even index equal to a negative number, that equation will have no solution.

If one side of an equation with a square root is a binomial, we use the Product of Binomial Squares Pattern when we square it.

Don’t forget the middle term!

Example 3

Solve: $\sqrt{p - 1} + 1 = p .$


To isolate the radical, subtract 1 from both sides.
Simplify.
Square both sides of the equation.
Simplify, using the Product of Binomial Squares Pattern on the right. Then solve the new equation.
It is a quadratic equation, so get zero on one side.
Factor the right side.
Use the Zero Product Property.
Solve each equation.
Check the answers.

	The solutions are $p = 1, p = 2.$

When the index of the radical is 3, we cube both sides to remove the radical.

${(\sqrt[3]{a})}^{3} = a$

Example 4

Solve: $\sqrt[3]{5 x + 1} + 8 = 4 .$

	$\sqrt[3]{5 x + 1} + 8 = 4$
To isolate the radical, subtract 8 from both sides.	$\sqrt[3]{5 x + 1} = −4$
Cube both sides of the equation.	${(\sqrt[3]{5 x + 1})}^{3} = {(−4)}^{3}$
Simplify.	$5 x + 1 = −64$
Solve the equation.	$5 x equals −65$
	$x equals −13$
Check the answer.

	The solution is $x equals −13.$

Sometimes an equation will contain rational exponents instead of a radical. We use the same techniques to solve the equation as when we have a radical. We raise each side of the equation to the power of the denominator of the rational exponent. Since ${(a^{m})}^{n} = a^{m \cdot n},$ we have for example,

${(x^{\frac{1}{2}})}^{2} = x, {(x^{\frac{1}{3}})}^{3} = x$

Remember, $x^{\frac{1}{2}} = \sqrt{x}$ and $x^{\frac{1}{3}} = \sqrt[3]{x} .$

Example 5

Solve: ${(3 x - 2)}^{\frac{1}{4}} + 3 = 5 .$

	${(3 x - 2)}^{\frac{1}{4}} + 3 = 5$
To isolate the term with the rational exponent, subtract 3 from both sides.	${(3 x - 2)}^{\frac{1}{4}} = 2$
Raise each side of the equation to the fourth power.	${({(3 x - 2)}^{\frac{1}{4}})}^{4} = {(2)}^{4}$
Simplify.	$3 x - 2 = 16$
Solve the equation.	$3 x equals 18$
	$x equals 6$
Check the answer.

	The solution is $x equals 6.$

Sometimes the solution of a radical equation results in two algebraic solutions, but one of them may be an extraneous solution!

Example 6

Solve: $\sqrt{r + 4} - r + 2 = 0 .$

	$\sqrt{r + 4} - r + 2 = 0$
Isolate the radical.	$\sqrt{r + 4} = r - 2$
Square both sides of the equation.	${(\sqrt{r + 4})}^{2} = {(r - 2)}^{2}$
Simplify and then solve the equation	$r + 4 = r^{2} - 4 r + 4$
It is a quadratic equation, so get zero on one side.	$0 = r^{2} - 5 r$
Factor the right side.	$0 = r (r - 5)$
Use the Zero Product Property.	$0 = r 0 = r - 5$
Solve the equation.	$r = 0 r = 5$
Check your answer.
	The solution is r = 5.
	$r equals 0$ is an extraneous solution.

When there is a coefficient in front of the radical, we must raise it to the power of the index, too.

Example 7

Solve: $3 \sqrt{3 x - 5} - 8 = 4 .$

	$3 \sqrt{3 x - 5} - 8 = 4$
Isolate the radical term.	$3 \sqrt{3 x - 5} = 12$
Isolate the radical by dividing both sides by 3.	$\sqrt{3 x - 5} = 4$
Square both sides of the equation.	${(\sqrt{3 x - 5})}^{2} = {(4)}^{2}$
Simplify, then solve the new equation.	$3 x - 5 = 16$
	$3 x equals 21$
Solve the equation.	$x equals 7$
Check the answer.

	The solution is $x equals 7.$

Solve Radical Equations with Two Radicals

If the radical equation has two radicals, we start out by isolating one of them. It often works out easiest to isolate the more complicated radical first.

In the next example, when one radical is isolated, the second radical is also isolated.

Example 8

Solve: $\sqrt[3]{4 x - 3} = \sqrt[3]{3 x + 2} .$

The radical terms are isolated.	$\sqrt[3]{4 x - 3} = \sqrt[3]{3 x + 2}$
Since the index is 3, cube both sides of the equation.	${(\sqrt[3]{4 x - 3})}^{3} = {(\sqrt[3]{3 x + 2})}^{3}$
Simplify, then solve the new equation.	$4 x - 3 = 3 x + 2$
	$x - 3 = 2$
	$x equals 5$
	$The solution is x = 5 .$
Check the answer.
We leave it to you to show that 5 checks!

Sometimes after raising both sides of an equation to a power, we still have a variable inside a radical. When that happens, we repeat Step 1 and Step 2 of our procedure. We isolate the radical and raise both sides of the equation to the power of the index again.

We summarize the steps here. We have adjusted our previous steps to include more than one radical in the equation This procedure will now work for any radical equations.

Be careful as you square binomials in the next example. Remember the pattern is ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ or ${(a - b)}^{2} = a^{2} - 2 a b + b^{2} .$

Example 10

Solve: $\sqrt{q - 2} + 3 = \sqrt{4 q + 1} .$


The radical on the right is isolated. Square both sides.
Simplify.
There is still a radical in the equation so we must repeat the previous steps. Isolate the radical.
Square both sides. It would not help to divide both sides by 6. Remember to square both the 6 and the $\sqrt{q - 2} .$
Simplify, then solve the new equation.
Distribute.
It is a quadratic equation, so get zero on one side.
Factor the right side.
Use the Zero Product Property.
The checks are left to you.	The solutions are $q equals 6$ and $q equals 2.$

Use Radicals in Applications

As you progress through your college courses, you’ll encounter formulas that include radicals in many disciplines. We will modify our Problem Solving Strategy for Geometry Applications slightly to give us a plan for solving applications with formulas from any discipline.

One application of radicals has to do with the effect of gravity on falling objects. The formula allows us to determine how long it will take a fallen object to hit the gound.

For example, if an object is dropped from a height of 64 feet, we can find the time it takes to reach the ground by substituting $h equals 64$ into the formula.



Take the square root of 64.
Simplify the fraction.

It would take 2 seconds for an object dropped from a height of 64 feet to reach the ground.

Example 11

Marissa dropped her sunglasses from a bridge 400 feet above a river. Use the formula $t = \frac{\sqrt{h}}{4}$ to find how many seconds it took for the sunglasses to reach the river.

Step 1. Read the problem.
Step 2. Identify what we are looking for.	the time it takes for the sunglasses to reach the river
Step 3. Name what we are looking.	Let $t =$ time.
Step 4. Translate into an equation by writing the appropriate formula. Substitute in the given information.
Step 5. Solve the equation.	$A mathematical expression reads 't = 20/4', indicating a variable 't' is equal to the fraction 20 divided by 4.$

Step 6. Check the answer in the problem and make sure it makes sense.
Does 5 seconds seem like a reasonable length of time?	Yes.
Step 7. Answer the question.	It will take 5 seconds for the sunglasses to reach the river.

Police officers investigating car accidents measure the length of the skid marks on the pavement. Then they use square roots to determine the speed, in miles per hour, a car was going before applying the brakes.

Example 12

After a car accident, the skid marks for one car measured 190 feet. Use the formula $s = \sqrt{24 d}$ to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Step 1. Read the problem
Step 2. Identify what we are looking for.	the speed of a car
Step 3. Name what weare looking for,	Let $s =$ the speed.
Step 4. Translate into an equation by writing the appropriate formula. Substitute in the given information.
Step 5. Solve the equation.

Round to 1 decimal place.

	The speed of the car before the brakes were applied was 67.5 miles per hour.

Key Concepts

Binomial Squares
$\begin{matrix} {(a + b)}^{2} = a^{2} + 2 a b + b^{2} \\ {(a - b)}^{2} = a^{2} - 2 a b + b^{2} \end{matrix}$
Solve a Radical Equation
1. Isolate one of the radical terms on one side of the equation.
2. Raise both sides of the equation to the power of the index.
3. Are there any more radicals?
  If yes, repeat Step 1 and Step 2 again.
  If no, solve the new equation.
4. Check the answer in the original equation.
Problem Solving Strategy for Applications with Formulas
1. Read the problem and make sure all the words and ideas are understood. When appropriate, draw a figure and label it with the given information.
2. Identify what we are looking for.
3. Name what we are looking for by choosing a variable to represent it.
4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.
Falling Objects
- On Earth, if an object is dropped from a height of h feet, the time in seconds it will take to reach the ground is found by using the formula $t = \frac{\sqrt{h}}{4} .$
Skid Marks and Speed of a Car
- If the length of the skid marks is d feet, then the speed, s, of the car before the brakes were applied can be found by using the formula $s = \sqrt{24 d} .$

Practice Makes Perfect

Solve Radical Equations

In the following exercises, solve.

$\sqrt{5 x - 6} = 8$

$x equals 14$

$\sqrt{4 x - 3} = 7$

$\sqrt{5 x + 1} = −3$

no solution

$\sqrt{3 y - 4} = −2$

$\sqrt[3]{2 x} = −2$

$x equals −4$

$\sqrt[3]{4 x - 1} = 3$

$\sqrt{2 m - 3} - 5 = 0$

$m equals 14$

$\sqrt{2 n - 1} - 3 = 0$

$\sqrt{6 v - 2} - 10 = 0$

$v equals 17$

$\sqrt{12 u + 1} - 11 = 0$

$\sqrt{4 m + 2} + 2 = 6$

$m = \frac{7}{2}$

$\sqrt{6 n + 1} + 4 = 8$

$\sqrt{2 u - 3} + 2 = 0$

no solution

$\sqrt{5 v - 2} + 5 = 0$

$\sqrt{u - 3} + 3 = u$

$u = 3, u = 4$

$\sqrt{v - 10} + 10 = v$

$\sqrt{r - 1} = r - 1$

$r = 1, r = 2$

$\sqrt{s - 8} = s - 8$

$\sqrt[3]{6 x + 4} = 4$

$x equals 10$

$\sqrt[3]{11 x + 4} = 5$

$\sqrt[3]{4 x + 5} - 2 = −5$

$x equals −8$

$\sqrt[3]{9 x - 1} - 1 = −5$

${(6 x + 1)}^{\frac{1}{2}} - 3 = 4$

$x equals 8$

${(3 x - 2)}^{\frac{1}{2}} + 1 = 6$

${(8 x + 5)}^{\frac{1}{3}} + 2 = −1$

$x equals −4$

${(12 x - 5)}^{\frac{1}{3}} + 8 = 3$

${(12 x - 3)}^{\frac{1}{4}} - 5 = −2$

$x equals 7$

${(5 x - 4)}^{\frac{1}{4}} + 7 = 9$

$\sqrt{x + 1} - x + 1 = 0$

$x equals 3$

$\sqrt{y + 4} - y + 2 = 0$

$\sqrt{z + 100} - z = −10$

$z equals 21$

$\sqrt{w + 25} - w = −5$

$3 \sqrt{2 x - 3} - 20 = 7$

$x equals 42$

$2 \sqrt{5 x + 1} - 8 = 0$

$2 \sqrt{8 r + 1} - 8 = 2$

$r equals 3$

$3 \sqrt{7 y + 1} - 10 = 8$

Solve Radical Equations with Two Radicals

In the following exercises, solve.

$\sqrt{3 u + 7} = \sqrt{5 u + 1}$

$u equals 3$

$\sqrt{4 v + 1} = \sqrt{3 v + 3}$

$\sqrt{8 + 2 r} = \sqrt{3 r + 10}$

$r equals −2$

$\sqrt{10 + 2 c} = \sqrt{4 c + 16}$

$\sqrt[3]{5 x - 1} = \sqrt[3]{x + 3}$

$x equals 1$

$\sqrt[3]{8 x - 5} = \sqrt[3]{3 x + 5}$

$\sqrt[3]{2 x^{2} + 9 x - 18} = \sqrt[3]{x^{2} + 3 x - 2}$

$x = −8, x = 2$

$\sqrt[3]{x^{2} - x + 18} = \sqrt[3]{2 x^{2} - 3 x - 6}$

$\sqrt{a} + 2 = \sqrt{a + 4}$

$a equals 0$

$\sqrt{r} + 6 = \sqrt{r + 8}$

$\sqrt{u} + 1 = \sqrt{u + 4}$

$u = \frac{9}{4}$

$\sqrt{x} + 1 = \sqrt{x + 2}$

$\sqrt{a + 5} - \sqrt{a} = 1$

$a equals 4$

$−2 = \sqrt{d - 20} - \sqrt{d}$

$\sqrt{2 x + 1} = 1 + \sqrt{x}$

$x = 0 x = 4$

$\sqrt{3 x + 1} = 1 + \sqrt{2 x - 1}$

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

$x = 1 x = 5$

$\sqrt{x + 1} - \sqrt{x - 2} = 1$

$\sqrt{x + 7} - \sqrt{x - 5} = 2$

$x equals 9$

$\sqrt{x + 5} - \sqrt{x - 3} = 2$

Use Radicals in Applications

In the following exercises, solve. Round approximations to one decimal place.

Landscaping Reed wants to have a square garden plot in his backyard. He has enough compost to cover an area of 75 square feet. Use the formula $s = \sqrt{A}$ to find the length of each side of his garden. Round your answer to the nearest tenth of a foot.

$8.7$ feet

Landscaping Vince wants to make a square patio in his yard. He has enough concrete to pave an area of 130 square feet. Use the formula $s = \sqrt{A}$ to find the length of each side of his patio. Round your answer to the nearest tenth of a foot.

Gravity A hang glider dropped his cell phone from a height of 350 feet. Use the formula $t = \frac{\sqrt{h}}{4}$ to find how many seconds it took for the cell phone to reach the ground.

$4.7$ seconds

Gravity A construction worker dropped a hammer while building the Grand Canyon skywalk, 4000 feet above the Colorado River. Use the formula $t = \frac{\sqrt{h}}{4}$ to find how many seconds it took for the hammer to reach the river.

Accident investigation The skid marks for a car involved in an accident measured 216 feet. Use the formula $s = \sqrt{24 d}$ to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

72 feet

Accident investigation An accident investigator measured the skid marks of one of the vehicles involved in an accident. The length of the skid marks was 175 feet. Use the formula $s = \sqrt{24 d}$ to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Writing Exercises

Explain why an equation of the form $\sqrt{x} + 1 = 0$ has no solution.

Answers will vary.

ⓐ Solve the equation $\sqrt{r + 4} - r + 2 = 0 .$ ⓑ Explain why one of the “solutions” that was found was not actually a solution to the equation.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

The table has 4 columns and 4 rows. The first row is a header row with the headers “I can…”, “Confidently”, “With some help.”, and “No – I don’t get it!”. The first column contains the phrases “Solve radical equations”, “solve radical equations with two radicals”, and “use radicals in applications”. The other columns are left blank so the learner can indicate their level of understanding.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?