Solve Radical Equations

By the end of this section, you will be able to: solve radical equations, solve radical equations with two radicals, use radicals in applications.

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.

Radical Equation

An equation in which a variable is in the radicand of a radical expression 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 nth power. This will eliminate the radical.

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

Example.

How to Solve a Radical Equation

Solve: $\sqrt{5 n - 4} - 9 = 0.$

Step	Result
1. Isolate the radical by adding 9 to both sides.	$\sqrt{5n-4}-9+9=0+9$
Simplify.	$\sqrt{5n-4}=9$
2. Square both sides.	$\left(\sqrt{5n-4}\right)^2=9^2$
Simplify.	$5n-4=81$
3. Solve the new equation.	$5n=85$
	$n=17$
4. Check the answer in the original equation.	$\sqrt{5(17)-4}-9=\sqrt{81}-9=9-9=0$

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

Solve: $\sqrt{10 z + 1} - 2 = 0$

How To

Solve a radical equation with one radical.

Step 1. Isolate the radical on one side of the equation.
Step 2. Raise both sides of the equation to the power of the index.
Step 3. Solve the new equation.
Step 4. Check the answer in the original equation.

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.

Example.

Solve: $\sqrt{9 k - 2} + 1 = 0.$

Step	Result
	$\sqrt{9k-2}+1=0$
To isolate the radical, subtract 1 from both sides.	$\sqrt{9k-2}+1-1=0-1$
Simplify.	$\sqrt{9k-2}=-1$

Because the square root is equal to a negative number, the equation has no solution.

Solve: $\sqrt{2 r - 3} + 5 = 0$

Solve: $\sqrt{7 s - 3} + 2 = 0$

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.

Binomial Squares

\begin{matrix} \\ \\ \left(a + b\right)^{2} = a^{2} + 2 a b + b^{2} \\ \left(a - b\right)^{2} = a^{2} - 2 a b + b^{2}\end{matrix}

Don’t forget the middle term!

Example.

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

Step	Result
	$\sqrt{p-1}+1=p$
To isolate the radical, subtract 1 from both sides.	$\sqrt{p-1}+1-1=p-1$
Simplify.	$\sqrt{p-1}=p-1$
Square both sides of the equation.	$\left(\sqrt{p-1}\right)^2=(p-1)^2$
Simplify, using the Product of Binomial Squares Pattern on the right. Then solve the new equation.	$p-1=p^2-2p+1$
It is a quadratic equation, so get zero on one side.	$0=p^2-3p+2$
Factor the right side.	$0=(p-1)(p-2)$
Use the Zero Product Property.	$0=p-1\quad\text{or}\quad 0=p-2$
Solve each equation.	$p=1\quad\text{or}\quad p=2$
Check $p=1$ .	$\sqrt{1-1}+1=1$
Check $p=2$ .	$\sqrt{2-1}+1=2$
	The solutions are $p = 1, p = 2.$

Solve: $\sqrt{x - 2} + 2 = x$

Solve: $\sqrt{y - 5} + 5 = y$

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

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

Example.

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

Step	Result
	$\sqrt[3]{5 x + 1} + 8 = 4$
To isolate the radical, subtract 8 from both sides.	$\sqrt[3]{5 x + 1} = -4$
Cube both sides of the equation.	$\left(\sqrt[3]{5 x + 1}\right)^{3} = \left(-4\right)^{3}$
Simplify.	$5 x + 1 = -64$
Solve the equation.	$5 x = -65$
	$x = -13$
Check the answer.	$\sqrt[3]{5(-13)+1}+8=\sqrt[3]{-64}+8=-4+8=4$
	The solution is $x = -13.$

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

Solve: $\sqrt[3]{6 x - 10} + 1 = -3$

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 $\left(a^{m}\right)^{n} = a^{m \cdot n},$ we have for example,

\left(x^{\tfrac{1}{2}}\right)^{2} = x, \left(x^{\tfrac{1}{3}}\right)^{3} = x

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

Example.

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

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

Solve: $\left(9 x + 9\right)^{\tfrac{1}{4}} - 2 = 1$

Solve: $\left(4 x - 8\right)^{\tfrac{1}{4}} + 5 = 7$

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

Example.

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

Step	Result
	$\sqrt{r + 4} - r + 2 = 0$
Isolate the radical.	$\sqrt{r + 4} = r - 2$
Square both sides of the equation.	$\left(\sqrt{r + 4}\right)^{2} = \left(r - 2\right)^{2}$
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.	$0 = r \left(r - 5\right)$
Use the Zero Product Property.	$0=r\quad\text{or}\quad0=r-5$
Solve the equation.	$r=0\quad\text{or}\quad r=5$
Check $r=0$ .	$\sqrt{0+4}-0+2=4\neq0,$ so $r=0$ is extraneous.
Check $r=5$ .	$\sqrt{5+4}-5+2=0,$ so the solution is $r=5.$

Solve: $\sqrt{m + 9} - m + 3 = 0$

Solve: $\sqrt{n + 1} - n + 1 = 0$

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

Example.

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

Step	Result
	$\text{3}\ \sqrt{3 x - 5} - 8 = 4$
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.	$\left(\sqrt{3 x - 5}\right)^{2} = \left(4\right)^{2}$
Simplify, then solve the new equation.	$3 x - 5 = 16$
	$3 x = 21$
Solve the equation.	$x = 7$
Check the answer.	$3\sqrt{3(7)-5}-8=3\sqrt{16}-8=12-8=4$
	The solution is $x = 7.$

Solve: $2 \sqrt{4 a + 4} - 16 = 16$

Solve: $3 \sqrt{2 b + 3} - 25 = 50$

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.

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

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

Solve: $\sqrt[3]{5 x - 4} = \sqrt[3]{2 x + 5}$

Solve: $\sqrt[3]{7 x + 1} = \sqrt[3]{2 x - 5}$

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.

Example.

How to Solve a Radical Equation

Solve: $\sqrt{m} + 1 = \sqrt{m + 9}.$

Step	Result
1. Isolate one radical. The radical on the right is already isolated.	$\sqrt{m}+1=\sqrt{m+9}$
2. Square both sides.	$\left(\sqrt{m}+1\right)^2=\left(\sqrt{m+9}\right)^2$
Simplify carefully.	$m+2\sqrt{m}+1=m+9$
3. Isolate the remaining radical.	$2\sqrt{m}=8$
Divide both sides by 2.	$\sqrt{m}=4$
Square both sides again.	$\left(\sqrt{m}\right)^2=4^2$
Solve.	$m=16$
4. Check the answer.	$\sqrt{16}+1=\sqrt{16+9}$
Simplify.	$4+1=5$

Solve: $3 - \sqrt{x} = \sqrt{x - 3}$

Solve: $\sqrt{x} + 2 = \sqrt{x + 16}$

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.

How To

Solve a radical equation.

Step 1. Isolate one of the radical terms on one side of the equation.
Step 2. Raise both sides of the equation to the power of the index.
Step 3. Are there any more radicals? If yes, repeat Step 1 and Step 2 again. If no, solve the new equation.
Step 4. Check the answer in the original equation.

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

Example.

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

Step	Result
	$\sqrt{q-2}+3=\sqrt{4q+1}$
The radical on the right is isolated. Square both sides.	$\left(\sqrt{q-2}+3\right)^2=\left(\sqrt{4q+1}\right)^2$
Simplify.	$q-2+6\sqrt{q-2}+9=4q+1$
There is still a radical in the equation, so repeat the previous steps. Isolate the radical.	$6\sqrt{q-2}=3q-6$
Square both sides. Remember to square both the 6 and $\sqrt{q-2}.$	$\left(6\sqrt{q-2}\right)^2=(3q-6)^2$
Expand the squares.	$6^2\left(\sqrt{q-2}\right)^2=(3q)^2-2(3q)(6)+6^2$
Simplify, then solve the new equation.	$36(q-2)=9q^2-36q+36$
Distribute.	$36q-72=9q^2-36q+36$
It is a quadratic equation, so get zero on one side.	$0=9q^2-72q+108$
Factor the right side.	$0=9(q-6)(q-2)$
Use the Zero Product Property.	$q-6=0\quad\text{or}\quad q-2=0$
Solve each equation.	$q=6\quad\text{or}\quad q=2$
The checks are left to you.	The solutions are $q = 6$ and $q = 2.$

Solve: $\sqrt{x - 1} + 2 = \sqrt{2 x + 6}$

Solve: $\sqrt{x} + 2 = \sqrt{3 x + 4}$

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.

How To

Use a problem solving strategy for applications with formulas.

Step 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.
Step 2. Identify what we are looking for.
Step 3. Name what we are looking for by choosing a variable to represent it.
Step 4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
Step 5. Solve the equation using good algebra techniques.
Step 6. Check the answer in the problem and make sure it makes sense.
Step 7. Answer the question with a complete sentence.

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.

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 = \tfrac{\sqrt{h}}{4}.

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 $h = 64$ into the formula.

Step	Result
	$t=\tfrac{\sqrt{h}}{4}$
Substitute $h=64$ .	$t=\tfrac{\sqrt{64}}{4}$
Take the square root of 64.	$t=\tfrac{8}{4}$
Simplify the fraction.	$t=2$

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

Example.

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

Step	Result
Step 1. Read the problem.
Step 2. Identify what we are looking for.	the time it takes for thesunglasses 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.	$t=\tfrac{\sqrt{400}}{4}$
Step 5. Solve the equation.	$t=\tfrac{20}{4}$
	$t=5$
Step 6. Check the answer in the problem and make sure it makes sense.	$5=\tfrac{\sqrt{400}}{4}=\tfrac{20}{4}=5$
Does 5 seconds seem like a reasonable length oftime?	Yes.
Step 7. Answer the question.	It will take 5 seconds for thesunglasses to reach the river.

A helicopter dropped a rescue package from a height of 1,296 feet. Use the formula $t = \tfrac{\sqrt{h}}{4}$ to find how many seconds it took for the package to reach the ground.

A window washer dropped a squeegee from a platform 196 feet above the sidewalk Use the formula $t = \tfrac{\sqrt{h}}{4}$ to find how many seconds it took for the squeegee to reach the sidewalk.

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.

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}

Example.

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	Result
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.	$s=\sqrt{24(190)}$
Step 5. Solve the equation.	$s=\sqrt{4{,}560}$
	$s=67.52777\ldots$
Round to 1 decimal place.	$s\approx67.5$
Step 6. Check the answer.	$67.5\approx\sqrt{24(190)}=\sqrt{4{,}560}\approx67.5277$
	The speed of the car before the brakes were appliedwas 67.5 miles per hour.

An accident investigator measured the skid marks of the car. The length of the skid marks was 76 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.

The skid marks of a vehicle involved in an accident were 122 feet long. 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.

This section is adapted from Intermediate Algebra 2e, Section 8.6: Solve Radical Equations by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: reformatted worked solutions for the web; omitted the Be Prepared quiz, media links, self-check reflection, and end-of-section exercise bank; and converted the source Try It practice into interactive exercises.

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