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Section 9.5 Equations with Radicals

Subsection Radical Equations

When an object falls from a height of \(h\) meters, the time it takes to reach the ground is given in seconds by

\begin{equation*} t=\sqrt{\dfrac{h}{4.9}} \end{equation*}

This equation is called a radical equation, because an unknown value, \(h\text{,}\) appears under a radical sign.

Radical Equation.

A radical equation is one in which the variable appears under a radical.

Example 9.44.

If you drop a penny from the top of the Sears Tower in Chicago, it will take 9.5 seconds to reach the ground. Find the height of the Sears Tower in meters by solving the equation

\begin{equation*} 9.5=\sqrt{\dfrac{h}{4.9}} \end{equation*}
Solution.

To solve the equation, we undo the operations performed on \(h\text{.}\) To undo the operation of taking a square root, we can square both sides of the equation. Remember that we undo the operations in the opposite order, as follows.

Operations performed on \(h\) \(\hphantom{0000}\) Steps for solution
1. Divide by 4.9 \(\hphantom{0000}\) 1. Square both sides
2. Take square root \(\hphantom{0000}\) 2. Multiply by 4.9

Here is the solution.

\begin{align*} 9.5^2 \amp = (\sqrt{\dfrac{h}{4.9}})^2 \amp \amp \blert{\text{Square both sides of the equation.}}\\ 90.25 \amp = \dfrac{h}{4.9} \amp \amp \blert{\text{Multiply both sides by 4.9.}}\\ 442.225 \amp = h \end{align*}

The Sears Tower is about 442 meters tall.

To Solve a Radical Equation.

  1. Isolate the radical on one side of the equation.

  2. Square both sides of the equation.

  3. Continue as usual to solve for the variable.

Reading Questions Reading Questions

1.

What is a radical equation?

Answer.

One in which the variable appears under a radical.

Example 9.45.

Solve the radical equation \(~~\sqrt{x-3} = 4\)

Solution.

We square both sides of the equation to produce an equation without radicals.

\begin{align*} (\sqrt{x-3})^2 \amp = 4^2\\ x-3 \amp = 16\\ x \amp = 19 \end{align*}

You can check that \(x=19\) satisfies the original equation.

Subsection Extraneous Solutions

The technique of squaring both sides may introduce extraneous solutions. (Recall that an extraneous solution is a value that is not a solution to the original equation.) For example, consider the equation

\begin{equation*} \sqrt{x+2}=-3 \end{equation*}

Squaring both sides gives us

\begin{align*} (\sqrt{x+2})^2 \amp =(-3)^2\\ x+2 \amp = 9\\ x \amp = 7 \end{align*}

However, if we substitute into the original equation, we see that it is not a solution.

\begin{equation*} \sqrt{\blert{7}+2} = \sqrt{9} = 3 \not= -3 \end{equation*}

The value \(x=7\) is a solution to the squared equation, but not to the original equation. In this case, the original equation has no solution.

Reading Questions Reading Questions

2.

When we solve a radical equation, what causes extraneous solutions to be introduced?

Answer.

Squaring both sides of the equation.

Extraneous Solutions.

Whenever we square both sides of an equation, we must check the solutions in the original equation.

If a radical equation involves several terms, it is easiest to isolate the radical term on one side of the equation before squaring both sides.

Example 9.46.

Solve \(~~4+\sqrt{8-2x} = x\)

Solution.

We first isolate the radical by subtracting from both sides to get

\begin{equation*} \sqrt{8-2x} = x-4 \end{equation*}

Now we square both sides to obtain

\begin{align*} (\sqrt{8-2x})^2 \amp = (x-4)^2\\ 8-2x \amp =x^2-8x+16 \amp \amp \blert{\text{Write the quadratic equation in standard form.}}\\ 0 \amp = x^2-6x+8 \amp \amp \blert{\text{Factor the right side.}}\\ 0 \amp = (x-4)(x-2) \end{align*}

The possible solutions are \(x=4\) and \(x=2\text{.}\) We check both of these in the original equation.

\(\blert{\text{Check:}}\) For \(x=\blert{4}\text{,}\)

\begin{align*} 4+\sqrt{8-2(\blert{4})} \amp = \blert{4}?\\ 4+\sqrt{0} \amp = 4?\\ 4 \amp = 4 \end{align*}

For \(x=\blert{2}\text{,}\)

\begin{align*} 4+\sqrt{8-2(\blert{2})} \amp = \blert{2}?\\ 4+\sqrt{4} \amp =2?\\ 6 \amp \not= 2 \end{align*}

Thus, \(x=2\) is an extraneous solution. The only solution to the original equation is \(x=4\text{.}\)

Caution 9.47.

When squaring both sides of an equation, we must be careful to square the entire expression on either side of the equal sign. It is incorrect to square each term separately. Thus, in Example 9.46, it would not be correct to write

\begin{equation*} ( \sqrt{8-2x})^2 = x^2-4^2 ~~~~~~~~~~~~\alert{\text{Incorrect!}} \end{equation*}

Reading Questions Reading Questions

3.

True or false: to solve a radical equation, we should square each term of the equation. Explain.

Answer.

False. We should square each side of the equation.

Subsection Equations with Cube Roots

We can also solve equations in which the variable appears under a cube root.

Example 9.48.

Solve the equation \(~~15-2\sqrt[3]{x-4} = 9\)

Solution.

We first isolate the cube root.

\begin{align*} 15-2\sqrt[3]{x-4} \amp = 9 \amp \amp \blert{\text{Subtract 15 from both sides.}}\\ -2\sqrt[3]{x-4} \amp = -6 \amp \amp \blert{\text{Divide both sides by}~-2.}\\ \sqrt[3]{x-4} \amp = 3 \end{align*}

Next, we undo the cube root by cubing both sides of the equation.

\begin{align*} (\sqrt[3]{x-4})^3 \amp = 3^3\\ x-4 \amp = 27 \end{align*}

Finally, we add 4 to both sides to find the solution, \(x=31\text{.}\) We do not have to check for extraneous solutions when we cube both sides of an equation, but it is a good idea to check the solution for accuracy anyway.

\(\blert{\text{Check:}}\) Substitute \(\alert{31}\) for \(x\) into the left side of the equation.

\begin{align*} 15-2\sqrt[3]{\alert{31}-4} \amp = 15-2\sqrt[3]{27} \\ \amp = 15-2(3)\\ \amp = 15-6=9 \amp \amp \blert{\text{The solution checks.}} \end{align*}

Reading Questions Reading Questions

4.

How can we solve an equation if the variable appears under a cube root?

Answer.

Isolate the cube root, then cube both sides.

Subsection Extraction of Roots

Now we'll compare solving radical equations, where we square both sides, with quadratic equations, where we may be able to take square roots of both sides.

Example 9.49.

Solve by extraction of roots \(~~2(3x-1)^2 = 36\)

Solution.

We first isolate the squared expression: we divide both sides by 2.

\begin{align*} (3x-1)^2 \amp = 18 \amp \amp \blert{\text{Extract square root.}}\\ 3x-1 \amp = \pm \sqrt{18} \amp \amp \blert{\text{Simplify the radical.}}\\ 3x-1 \amp = \pm 3\sqrt{2} \amp \amp \blert{\text{Write as two equations.}} \end{align*}

Now we solve each equation, to find two solutions.

\begin{align*} 3x-1\amp = 3\sqrt{2} \amp 3x-1\amp = -3\sqrt{2}\\ 3x \amp = 1 +3\sqrt{2} \amp 3x \amp = 1 -3\sqrt{2}\\ x \amp = \dfrac{1 +3\sqrt{2}}{2} \amp x \amp = \dfrac{1 -3\sqrt{2}}{2} \end{align*}

We write the solutions as \(~\dfrac{1 \pm 3\sqrt{2}}{2}\text{.}\)

Reading Questions Reading Questions

5.

Is it true that \(1+3\sqrt{2} = 4\sqrt{2}\text{?}\) Why or why not?

Answer.

No. \(1\) and \(3\sqrt{2}\) are not like terms.

Subsection Skills Warm-Up

Exercises Exercises

Exercise Group.

Square each expression.

1.

  1. \(\displaystyle 3x\)

  2. \(\displaystyle 3+x\)

2.

  1. \(\displaystyle \sqrt{3x}\)

  2. \(\displaystyle 3\sqrt{x}\)

3.

  1. \(\displaystyle \sqrt{x+3}\)

  2. \(\displaystyle \sqrt{x}+\sqrt{3}\)

4.

  1. \(\displaystyle x-\sqrt{3}\)

  2. \(\displaystyle 3-\sqrt{x}\)

5.

  1. \(\displaystyle 2-3\sqrt{x}\)

  2. \(\displaystyle 2\sqrt{2x-3}\)

6.

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

  2. \(\displaystyle 3+2\sqrt{x-3}\)

Subsubsection Answers to Skills Warm-Up

    1. \(\displaystyle 9x^2\)

    2. \(\displaystyle 9+6x+x^2\)

    1. \(\displaystyle 3x\)

    2. \(\displaystyle 9x\)

    1. \(\displaystyle x+3\)

    2. \(\displaystyle x+2\sqrt{3x}+3\)

    1. \(\displaystyle x^2-2\sqrt{3}x+3\)

    2. \(\displaystyle 9-6\sqrt{x}+x\)

    1. \(\displaystyle 4-12\sqrt{x}+9x\)

    2. \(\displaystyle 8x-24\)

    1. \(\displaystyle x=1+4\sqrt{x-3}\)

    2. \(\displaystyle 4x-3+12\sqrt{x-3}\)

Subsection

Subsubsection Activity 1: Solving Radical Equations

Exercises Exercises
Exercise Group.

After solving a radical equation, remember to check for extraneous solutions.

1.

Solve \(~\sqrt{x-6}=2\)

2.

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

3.

Solve \(~3\sqrt[3]{4x-1}=-15\)

Subsubsection Activity 2: Roller Coasters

You have been commissioned to design a new roller coaster for an amusement park. The roller coaster should be more spectacular than all existing roller coasters, and in particular it should include a vertical loop. In order to stay on the track through the loop, the cars must travel at a speed given (in miles per hour) by

\begin{equation*} v=\sqrt{89.3r} \end{equation*}

where \(r\) is the radius of the loop in feet.

Here are three roller coasters with ver tall vertical loops:

Height Roller Coaster Location
188 feet Viper Six Flags Magic Mountain, Valencia, California
173 feet Gash Six Flags Great Adventure, Jackson, New Jersey
170 feet Shockwave Six Flags Great Adventure, Gurnee, Illinois
Exercises Exercises
1.

Evaluate the formula to find the speed the cars must reach on each roller coaster. Because the vertical loops used in roller coasters are not perfect circles, the total height of the loop is about 2.5 times its radius.

Viper:

Gash:

Shockwave:

loop
2.

You would like your roller coaster to have a vertical loop that is 200 feet tall. How fast must the cars travel?

3.

Now suppose you know the maximum speed possible for the cars on a particular roller coaster. Can you calculate the height of the tallest vertical loop the cars can negotiate? Here are the speeds of three of the world's fastest roller coasters.

Speed Roller Coaster Location
86 mph Fujiyama Fujikyu Highland Park, Japan
80 mph Steel Phantom Kennywood, West Miflin, Pennsylvania
79 mph Desperado Buffalo Bill's, Jean, Nevada

Calculate the maximum loop height possible for each roller coaster.

Desperado:

Steel Phantom:

Fujiyama:

Subsubsection Activity 3: Quadratic Equations

Exercises Exercises
1.

Solve \(~~(2x+1)^2=8\)

2.

Solve \(~~3x^2-6x+2=0~~\) and simplify the solutions.

Subsubsection Wrap-Up

Objectives.

In this Lesson we practiced the following skills:

  • Solving radical equations

  • Checking for extraneous solutions

  • Simplifying solutions of quadratic equations

Questions.

  1. What operation might introduce extraneous solutions?

  2. What is wrong with this strategy: To solve an equation with a variable under a square root, square each term of the equation.

  3. How many solutions does the equation \(~x^2=64~\) have? How many solutions does \(~x^3=-64~\) have?

Subsection Homework Preview

Exercises Exercises

Exercise Group.

Solve.

1.

\(5\sqrt{2x-4}-7=23\)

2.

\(4\sqrt[3]{6-x}+5=-3\)

Exercise Group.

Solve, and check for extraneous solutions.

3.

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

4.

\(t-5=\sqrt{2t+5}\)

5.

Solve by extracting roots \(~~3(2x-3)^2=15\)

6.

Solve \(~~2x^2-4x-5=0~~\) and simplify your answer.

Subsubsection Answers to Homework Preview

  1. \(\displaystyle x=20\)

  2. \(\displaystyle x=14\)

  3. No solution

  4. \(\displaystyle t=10\)

  5. \(\displaystyle x=\dfrac{3 \pm \sqrt{5}}{2}\)

  6. \(\displaystyle x=\dfrac{2 \pm \sqrt{14}}{2}\)

Exercises Homework 9.5

Exercise Group.

For Problems 1–5, solve and check.

1.

\(\sqrt{x+4}=5\)

2.

\(\sqrt{x}-4=5\)

3.

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

4.

\(2+3\sqrt{x-1}=8\)

5.

\(2\sqrt{3x+1}-3=5\)

6.

Use the graph of \(y=\sqrt{x}\) shown below to solve the equations. (You may have to estimate some of the solutions.) Check your answers algebraically.

square root graph

  1. \(\displaystyle \sqrt{x} = 4\)

  2. \(\displaystyle \sqrt{x} = 2.5\)

  3. \(\displaystyle \sqrt{x} = -2\)

  4. \(\displaystyle \sqrt{x} = 5.3\)

7.

The equation for the semicircle shown is

\begin{equation*} y=\sqrt{9-x^2} \end{equation*}

Find the \(x\)-coordinates of two different points on the semicircle that have \(y\)-coordinate 2.

semicircle
Exercise Group.

For Problems 8–12, solve, and check for extraneous solutions.

8.

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

9.

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

10.

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

11.

\(\sqrt{x+7}=2x-4\)

12.

\(6+\sqrt{5x-4}-x=4\)

Exercise Group.

For Problems 13–15, solve.

13.

\(2\sqrt[3]{x}+15=5\)

14.

\(\sqrt[3]{2x-5}-1=2\)

15.

\(2=8-3\sqrt[3]{x^3+1}\)

Exercise Group.

For Problems 16–18, solve by extraction of roots, and simplify your answer.

16.

\((2x+1)^2=8\)

17.

\(3(2x-8)^2=60\)

18.

\(\dfrac{4}{3}(x+3)^2=24\)

Exercise Group.

For Problems 19–20, solve the formula for the indicated variable.

19.

\(x^2+a^2=b^2~~~~\) for \(x\)

20.

\(\dfrac{x^2}{4}-y^2=1~~~~\) for \(x\)

21.

  1. Complete the table of values and graph \(y=\sqrt{x-4}\text{.}\)

    \(x\) \(y\)
    \(4\) \(\hphantom{0000}\)
    \(5\) \(\hphantom{0000}\)
    \(6\) \(\hphantom{0000}\)
    \(10\) \(\hphantom{0000}\)
    \(16\) \(\hphantom{0000}\)
    \(19\) \(\hphantom{0000}\)
    \(24\) \(\hphantom{0000}\)
    grid

  2. Solve \(~\sqrt{x-4}=3~\) graphically and algebraically. Do your answers agree?

22.

  1. Complete the table of values and graph \(y=4-\sqrt{x+3}\text{.}\)

    \(x\) \(y\)
    \(-3\) \(\hphantom{0000}\)
    \(-2\) \(\hphantom{0000}\)
    \(0\) \(\hphantom{0000}\)
    \(1\) \(\hphantom{0000}\)
    \(4\) \(\hphantom{0000}\)
    \(8\) \(\hphantom{0000}\)
    \(16\) \(\hphantom{0000}\)
    grid

  2. Solve \(~4-\sqrt{x+3}=1~\) graphically and algebraically. Do your answers agree?

23.

The higher your altitude, the farther you can see to the horizon, if nothing blocks your line of sight. From a height of \(h\) meters, the distance \(d\) to the horizon in kilometers is given by

\begin{equation*} d=\sqrt{12h} \end{equation*}

  1. Mt. Wilson is part of the San Gabriel mountains north of Los Angeles, and it has an elevation of 1740 meters. How far can you see from Mt. Wilson?

  2. The new Getty Center is built on the hills above Sunset Boulevard in Los Angeles, and from the patio on a clear day you can see the city of Long Beach, 44 kilometers away. What does this tell you about the elevation at the Getty Center?

24.

The speed of a tsunami is given, in miles per hour, by

\begin{equation*} s=3.9\sqrt{d} \end{equation*}

where \(d\) is the depth of the ocean beneath the wave, in feet. A tsunami traveling along the Aleutian Trench off the coast of Alaska is moving at a speed of over 615 miles per hour. Find the depth of Aleutian Trench.

25.

The height of a cylindrical storage tank is four times its radius. If the tank holds \(V\) cubic inches of liquid, its radius in inches is

\begin{equation*} r=\sqrt[3]{\dfrac{V}{12.57}} \end{equation*}

  1. If the tank should hold 340 cubic inches, what must its radius be?

  2. If the radius of the tank is 5.5 inches, how much liquid can it hold?