## Radical Equation Word Problems - Examples & Practice - Expii

Radical equation word problems - examples & practice, explanations (3).

The key to solving any word problem (whether it contains a radical or not) is to translate the problem from words into math .

That's the biggest step in word problems. Once you've translated the information into numbers, you solve the equation the same way as always.

Let's look at an example to see how this approach works when radicals are involved.

Hiroki is calculating the speed of a tsunami. The formula is speed=√9.8d where d is ocean depth in meters. In this case, the ocean was 1500 meters deep. How fast was the wave (in m/s2)?

So, in this problem (as with many problems involving radicals), the formula to use has already been given to us. We are interested in the speed of the wave, and we've been given the depth (d). So, let's start by plugging that value in : speed=√9.8(1500) Let's solve this problem.

How fast was the wave (m/s2)?

## Related Lessons

Example walk through: word problem involving radicals.

The key to word problems is in the translation. Let's walk through the problem below and solve it.

Lori recently purchased a square painting with an area of 780 cm2. How much wood does Lori need to make the frame? Don't worry about the width and height of the wood, just find the length.

First, find out the essential information in the problem.

A square painting has area 780cm2, what is its side length?

What is the relationship between side length and area of a square? Area=side length×side length

Denote side length as x for simplicity, plug in the numbers we have: 780cm2=x×x Solve for x: 780=x2x=√78x≈27.93

Lastly, note that the question is asking how much wood Lori needs to build the frame. A square has 4 sides so we have to multiply 27.93cm by 4.

In total, Lori needs at least 27.93cm×4=111.72cm of wood.

Let's look at another problem!

Image source: By Caroline Kulczycky

## (Video) Problem Solving with Radical Expressions

by Deanna Green

This video by Deanna Green works through word problems with radicals.

These problems usually focus around the Pythagorean Theorem . A helpful trick always is to draw a simple picture of what's going on. Then fill in lengths or angles as the problem states.

Let's work through one of the examples she looks at.

A kite is secured to a rope that is tied to the ground. A breeze blows the kite so that the rope is taught while the kite is directly above a flagpole that is 30 ft from where the rope is staked down. Find the altitude of the kite if the rope is 110 ft long.

The first step is to draw a diagram.

Image source: by Alex Federspiel

The Pythagorean Theorem states:

The length of the hypotenuse (diagonal line) is equal to the square root of the square of the sides added.

As an equation:

where c is the hypotenuse and a and b are the other sides.

Start by plugging in all the information we know.

110=√302+x2

You cannot split up radicals with a sum underneath into two separate radicals. So, to solve this, we want to take the square of both sides.

(110)2=(√302+x2)212100=302+x212100−302=x212100−900=x211200=x2√11200=√x2105.83≈x

You may need to review the lesson on getting rid of the x2 in with this lesson .

## Learning Objectives

By the end of this section, you will be able to:

Before you get started, take this readiness quiz.

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 .

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 n th power. This will eliminate the radical.

• 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.

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.

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!

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

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

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

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.

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.

• Isolate one of the radical terms on one side of the equation.

If yes, repeat Step 1 and Step 2 again.

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.

• 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.
• Identify what we are looking for.
• Name what we are looking for by choosing a variable to represent it.
• Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
• Solve the equation using good algebra techniques.
• Check the answer in the problem and make sure it makes sense.
• 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.

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

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

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.

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

Access these online resources for additional instruction and practice with solving radical equations.

• Solving an Equation Involving a Single Radical
• Solving Equations with Radicals and Rational Exponents

## Practice Makes Perfect

In the following exercises, solve.

no solution

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

## Writing Exercises

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

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

How to solve equations with square roots, cube roots, etc.

We can get rid of a square root by squaring (or cube roots by cubing, etc).

Warning: this can sometimes create "solutions" which don't actually work when we put them into the original equation. So we need to Check!

• isolate the square root on one side of the equation
• square both sides of the equation

Then continue with our solution!

## Example: solve √(2x+9) − 5 = 0

Now it should be easier to solve!

Check: √(2·8+9) − 5 = √(25) − 5 = 5 − 5 = 0

That one worked perfectly.

## More Than One Square Root

What if there are two or more square roots? Easy! Just repeat the process for each one.

It will take longer (lots more steps) ... but nothing too hard.

## Example: solve √(2x−5) − √(x−1) = 1

We have removed one square root.

Now do the "square root" thing again:

We have now successfully removed both square roots.

Let us continue on with the solution.

It is a Quadratic Equation! So let us put it in standard form.

Using the Quadratic Formula (a=1, b=−14, c=29) gives the solutions:

2.53 and 11.47 (to 2 decimal places)

Let us check the solutions:

There is really only one solution :

Answer: 11.47 (to 2 decimal places)

See? This method can sometimes produce solutions that don't really work!

The root that seemed to work, but wasn't right when we checked it, is called an "Extraneous Root"

So checking is important.

## How To : Solve word problems containing radical equations

See how to unpack and solve a word problem containing radical equations with this free video math lesson from Internet pedagogical superstar Simon Khan. From Ramanujan to calculus co-creator Gottfried Leibniz, many of the world's best and brightest mathematical minds have belonged to autodidacts. And, thanks to the Internet, it's easier than ever to follow in their footsteps (or just finish your homework or study for that next big test).

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Step 1. Plug in the known g and h values. v = 2 ( 32) ( 180) Step 2. Simplify to find the unknown velocity v . v = 2 ( 32) ( 180) = 11520 = 107.33 The velocity of the pinecone is approximately 107...

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How to: Solve a Radical Equation. Isolate a radical. Put ONE radical on one side of the equal sign and put everything else on the other side. Eliminate the radical. Raise both sides of the equal sign to the power that matches the index on the radical. This means square both sides if it is a square root; cube both sides if it is a cube root; etc.

Solve: Solve: Solve: Solve a radical equation with one radical. 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. When we use a radical sign, it indicates the principal or positive root.

A Radical Equation is an equation with a square root or cube root, etc. Solving Radical Equations. We can get rid of a square root by squaring (or cube roots by cubing, etc). Warning: this can sometimes create "solutions" which don't actually work when we put them into the original equation. So we need to Check!

Key Steps to Solve Radical Equations: 1) Isolate the radical symbol on one side of the equation 2) Square both sides of the equation to eliminate the radical symbol 3) Solve the equation that comes out after the squaring process 4) Check your answers with the original equation to avoid extraneous values or solutions

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20. How to Solve word problems containing radical equations

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