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Solve Radical Equations

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.

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

How to Solve a Radical Equation

Try it.

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

Solution

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

Try it.

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

Solution


To isolate the radical, subtract 1 to both sides.
Simplify.

Because the square root is equal to a negative number, the equation has 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!

\[{(\sqrt[3]{a})}^{3}=a\]\[{({x}^{\frac{1}{2}})}^{2}=x,\ {({x}^{\frac{1}{3}})}^{3}=x\]

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

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

Try it.

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

Solution

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

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.

How to Solve a Radical Equation

Try it.

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

Solution

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 \({(a+b)}^{2}={a}^{2}+2ab+{b}^{2}\) or \({(a-b)}^{2}={a}^{2}-2ab+{b}^{2}.\)

Example

Try it.

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

Solution


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 \(q=6\) and \(q=2.\)

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 \(h=64\) 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

Try it.

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.

Solution

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.

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.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Key Concepts

Binomial Squares
\(\begin{array}{l}{(a+b)}^{2}={a}^{2}+2ab+{b}^{2} \\ {(a-b)}^{2}={a}^{2}-2ab+{b}^{2}\end{array}\)
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 \(s=\sqrt{24d}.\)

Solve Radical Equations

Solve Radical Equations

In the following exercises, solve.

Try it.

\(\sqrt{5x-6}=8\)

Solution

\(x=14\)

Try it.

\(\sqrt{4x-3}=7\)

Try it.

\(\sqrt{5x+1}=-3\)

Solution

no solution

Try it.

\(\sqrt{3y-4}=-2\)

Try it.

\(\sqrt[3]{2x}=-2\)

Solution

\(x=-4\)

Try it.

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

Try it.

\(\sqrt{2m-3}-5=0\)

Solution

\(m=14\)

Try it.

\(\sqrt{2n-1}-3=0\)

Try it.

\(\sqrt{6v-2}-10=0\)

Solution

\(v=17\)

Try it.

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

Try it.

\(\sqrt{4m+2}+2=6\)

Solution

\(m=\frac{7}{2}\)

Try it.

\(\sqrt{6n+1}+4=8\)

Try it.

\(\sqrt{2u-3}+2=0\)

Solution

no solution

Try it.

\(\sqrt{5v-2}+5=0\)

Try it.

\(\sqrt{u-3}+3=u\)

Solution

\(u=3,u=4\)

Try it.

\(\sqrt{v-10}+10=v\)

Try it.

\(\sqrt{r-1}=r-1\)

Solution

\(r=1,r=2\)

Try it.

\(\sqrt{s-8}=s-8\)

Try it.

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

Solution

\(x=10\)

Try it.

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

Try it.

\(\sqrt[3]{4x+5}-2=-5\)

Solution

\(x=-8\)

Try it.

\(\sqrt[3]{9x-1}-1=-5\)

Try it.

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

Solution

\(x=8\)

Try it.

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

Try it.

\({(8x+5)}^{\frac{1}{3}}+2=-1\)

Solution

\(x=-4\)

Try it.

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

Try it.

\({(12x-3)}^{\frac{1}{4}}-5=-2\)

Solution

\(x=7\)

Try it.

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

Try it.

\(\sqrt{x+1}-x+1=0\)

Solution

\(x=3\)

Try it.

\(\sqrt{y+4}-y+2=0\)

Try it.

\(\sqrt{z+100}-z=-10\)

Solution

\(z=21\)

Try it.

\(\sqrt{w+25}-w=-5\)

Try it.

\(3\sqrt{2x-3}-20=7\)

Solution

\(x=42\)

Try it.

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

Try it.

\(2\sqrt{8r+1}-8=2\)

Solution

\(r=3\)

Try it.

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

Solve Radical Equations with Two Radicals

In the following exercises, solve.

Try it.

\(\sqrt{3u+7}=\sqrt{5u+1}\)

Solution

\(u=3\)

Try it.

\(\sqrt{4v+1}=\sqrt{3v+3}\)

Try it.

\(\sqrt{8+2r}=\sqrt{3r+10}\)

Solution

\(r=-2\)

Try it.

\(\sqrt{10+2c}=\sqrt{4c+16}\)

Try it.

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

Solution

\(x=1\)

Try it.

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

Try it.

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

Solution

\(x=-8,x=2\)

Try it.

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

Try it.

\(\sqrt{a}+2=\sqrt{a+4}\)

Solution

\(a=0\)

Try it.

\(\sqrt{r}+6=\sqrt{r+8}\)

Try it.

\(\sqrt{u}+1=\sqrt{u+4}\)

Solution

\(u=\frac{9}{4}\)

Try it.

\(\sqrt{x}+1=\sqrt{x+2}\)

Try it.

\(\sqrt{a+5}-\sqrt{a}=1\)

Solution

\(a=4\)

Try it.

\(-2=\sqrt{d-20}-\sqrt{d}\)

Try it.

\(\sqrt{2x+1}=1+\sqrt{x}\)

Solution

\(x=0\ x=4\)

Try it.

\(\sqrt{3x+1}=1+\sqrt{2x-1}\)

Try it.

\(\sqrt{2x-1}-\sqrt{x-1}=1\)

Solution

\(x=1\ x=5\)

Try it.

\(\sqrt{x+1}-\sqrt{x-2}=1\)

Try it.

\(\sqrt{x+7}-\sqrt{x-5}=2\)

Solution

\(x=9\)

Try it.

\(\sqrt{x+5}-\sqrt{x-3}=2\)

Use Radicals in Applications

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

Try it.

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.

Solution

\(8.7\) feet

Try it.

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.

Try it.

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.

Solution

\(4.7\) seconds

Try it.

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.

Try it.

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

Solution

72 feet

Try it.

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 \(s=\sqrt{24d}\) to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Simplify: \({(y-3)}^{2}.\)
If you missed this problem, review .

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\({y}^{2}-6y+9\)
Solve: \(2x-5=0.\)
If you missed this problem, review .

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\(x=\frac{5}{2}\)
Solve \({n}^{2}-6n+8=0.\)
If you missed this problem, review .

Առաջարկել պատասխանը

\(n=2\ \text{or}\ n=4\)
Solve: \(\sqrt{5n-4}-9=0.\)

Առաջարկել պատասխանը
Solve: \(\sqrt{3m+2}-5=0.\)

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\(m=\frac{23}{3}\)
Solve: \(\sqrt{10z+1}-2=0.\)

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\(z=\frac{3}{10}\)
Solve: \(\sqrt{9k-2}+1=0.\)

Առաջարկել պատասխանը

To isolate the radical, subtract 1 to both sides.
Simplify.

Because the square root is equal to a negative number, the equation has no solution.
Solve: \(\sqrt{2r-3}+5=0.\)

Առաջարկել պատասխանը

\(\text{no solution}\)
Solve: \(\sqrt{7s-3}+2=0.\)

Առաջարկել պատասխանը

\(\text{no solution}\)

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

Առաջարկել պատասխանը


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 \(p=1,\ p=2.\)

Solve: \(\sqrt{x-2}+2=x.\)

Առաջարկել պատասխանը

\(x=2,x=3\)
Solve: \(\sqrt{y-5}+5=y.\)

Առաջարկել պատասխանը

\(y=5,y=6\)

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

Առաջարկել պատասխանը

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

	The solution is \(x=-13.\\)

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

Առաջարկել պատասխանը

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

Առաջարկել պատասխանը

\(x=-9\)

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

Առաջարկել պատասխանը

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

	The solution is \(x=6.\\)

Solve: \({(9x+9)}^{\frac{1}{4}}-2=1.\)

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\(x=8\)
Solve: \({(4x-8)}^{\frac{1}{4}}+5=7.\)

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\(x=6\)

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

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	\(\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}-4r+4\\)
It is a quadratic equation, so get zero on one side.	\(0={r}^{2}-5r\\)
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.
	\(r=0\) is an extraneous solution.

Solve: \(\sqrt{m+9}-m+3=0.\)

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\(m=7\)
Solve: \(\sqrt{n+1}-n+1=0.\)

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\(n=3\)

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

Առաջարկել պատասխանը

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

	The solution is \(x=7.\\)

Solve: \(2\sqrt{4a+4}-16=16.\)

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\(a=63\)
Solve: \(3\sqrt{2b+3}-25=50.\)

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\(b=311\)

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

Առաջարկել պատասխանը

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

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

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

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\(x=-\frac{6}{5}\)
Solve: \(\sqrt{m}+1=\sqrt{m+9}.\)

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Solve: \(3-\sqrt{x}=\sqrt{x-3}.\)

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\(x=4\)
Solve: \(\sqrt{x}+2=\sqrt{x+16}.\)

Առաջարկել պատասխանը

\(x=9\)

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

Առաջարկել պատասխանը


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 \(q=6\) and \(q=2.\)

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

Առաջարկել պատասխանը

\(x=5\)
Solve: \(\sqrt{x}+2=\sqrt{3x+4}\)

Առաջարկել պատասխանը

\(x=0\ x=4\)

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.

Առաջարկել պատասխանը

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.

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.

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

Առաջարկել պատասխանը

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

Առաջարկել պատասխանը

\(3.5\) seconds

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

Առաջարկել պատասխանը

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.

An accident investigator measured the skid marks of the car. The length of the skid marks was 76 feet. Use the formula \(s=\sqrt{24d}\) to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Առաջարկել պատասխանը

\(42.7\) feet
The skid marks of a vehicle involved in an accident were 122 feet long. Use the formula \(s=\sqrt{24d}\) to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Առաջարկել պատասխանը

\(54.1\) feet
\(\sqrt{5x-6}=8\)

Առաջարկել պատասխանը

\(x=14\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Solve Radical Equations

Solve radical equations
Solve radical equations with two radicals
Use radicals in applications
Isolate the radical on one side of the equation.
Raise both sides of the equation to the power of the index.
Solve the new equation.
Check the answer in the original equation.
Isolate one of the radical terms on one side of the equation.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Փորձեք ինքներդ

Parts of this page are adapted from OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Ցուցադրել Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value