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Section 6.2 Roots and Radicals

In the previous section we saw that inverse variation can be expressed as a power function by using negative integer exponents. For example, the function f(x)=kx2 can also be written as f(x)=kx2. Using negative exponents can simplify many calculations that would otherwise involve algebraic fractions.
But is there some meaning to exponents that are fractions, such as 12 or 13? For example, do expressions such as 51/2 or a1/3 have some familiar interpretation? We’ll see that fractional exponents can denote square roots and other radicals.

Subsection 6.2.1 nth Roots

Recall that s is a square root of b if s2=b, and s is a cube root of b if s3=b. For example, 5 is a square root of 25 because 52=25, and 2 is a cube root of 8 because 23=8.
In a similar way, we can define the fourth, fifth, or sixth root of a number. For instance, the fourth root of 81 is the number 3 because 34=81.

nth Roots.

s is called an nth root of b if sn=b.
We use the symbol bn to denote the nth root of b. An expression of the form bn is called a radical, b is called the radicand, and n is called the index of the radical.
 [TK]

Example 6.2.1.

  1. 814=3 because 34=81
  2. 325=2 because 25=32
  3. 646=2 because 26=64
  4. 14=1 because 14=1
  5. 100,0005=10 because 105=100,000

Checkpoint 6.2.2. Practice 1.

Evaluate each radical.
  1. 164
  2. 2435
Solution.
  1. 2
  2. 3

Subsection 6.2.2 Exponential Notation for Radicals

A convenient notation for radicals uses fractional exponents. Consider the expression 91/2. What meaning can we attach to an exponent that is a fraction? Recall that when we raise a power to a power, we multiply the exponents together:
(xa)b=xab
Therefore, if we square the number 91/2, we get
(91/2)2=9(1/2)(2)=91=9
Thus, 91/2 is a number whose square is 9. But this means that 91/2 is a square root of 9, or
91/2=9=3

Checkpoint 6.2.3. QuickCheck 1.

Fill in the blanks.
  1. Inverse variation can be expressed as a power function by using a exponent.
  2. The number n in the expression bn is called the of the radical.
  3. We use exponents to denote radicals.
  4. State the law of exponents that shows that a1/2 is a square root of a.
Remember that 9 actually has two square roots, namely 3 and 3. But the symbol 00 represents only the positive square root, so that 9=3. The same is true for the exponent 12: any nonnegative number raised to the 12 power is equal to the positive square root of the number, or
a1/2=a

Example 6.2.4.

  1. 251/2=5
  2. 251/2=5
  3. (25)1/2 is not a real number.
  4. 01/2=0

Checkpoint 6.2.5. Practice 2.

Evaluate each power.
  1. 41/2
  2. 42
  3. 41/2
  4. (14)1/2
Solution.
  1. 2
  2. 116
  3. 12
  4. 12
The same reasoning works for roots with any index. For instance, 81/3 is the cube root of 8, because
(81/3)3=8(1/3)(3)=81=8
Thus, we make the following definition for fractional exponents.

Exponential Notation for Radicals.

For any integer n2 and for a0,
a1/n=an

Example 6.2.6.

  1. 811/4=814=3
  2. 1251/3=1253=5

Caution 6.2.7.

Note that
251/212(25)   and   1251/313(125)
An exponent of 12 denotes the square root of its base, and an exponent of 13 denotes the cube root of its base.

Checkpoint 6.2.8. Practice 3.

Write each power with radical notation, and then evaluate.
  1. 321/5
  2. 6251/4
Solution.
  1. 325=2
  2. 6254=5
Of course, we can use decimal fractions for exponents as well. For example,
a=a1/2=a0.5     and     a4=a1/4=a0.25

Example 6.2.9.

  1. 1000.5=100=10
  2. 160.25=164=2

Checkpoint 6.2.10. Practice 4.

Write each power with radical notation, and then evaluate.
  1. 100,0000.2
  2. 810.25
Solution.
  1. 100,0005=10
  2. 814=3

Checkpoint 6.2.11. QuickCheck 2.

True or false.
  1. The exponent 1n tells us to take a reciprocal.
  2. The exponent 0.2 denotes a fifth root.
  3. A power with a negative exponent is not a real number.
  4. The expressions x3 and x1/3 are the same.

Subsection 6.2.3 Irrational Numbers

What about nth roots such as 23 and 51/3 that cannot be evaluated easily? These are examples of irrational numbers. An irrational number is one that cannot be expressed as a quotient of two integers.
It is not possible to write down an exact decimal equivalent for an irrational number, but we can find an approximation to as many decimal places as we like.  [TK]  We can use a calculator to obtain decimal approximations for irrational numbers. For example, you can verify that
234.796      and      51/31.710

Caution 6.2.12.

The following keying sequence for evaluating the irrational number 71/5 is incorrect:
7 ^ 1 ÷ 5 ENTER
You can check that this sequence calculates 715, instead of 71/5. Recall that according to the order of operations, powers are computed before multiplications or divisions. We must enclose the exponent 1/5 in parentheses and enter
7 ^ ( 1 ÷ 5 ) ENTER
Or, because 15=0.2, we can enter
7 ^ 0.2 ENTER

Subsection 6.2.4 Working with Fractional Exponents

Fractional exponents simplify many calculations involving radicals. It will be useful for you to convert easily between exponential and radical notation.

Example 6.2.13.

Convert each radical to exponential notation.
  1. 123=121/3
  2. 2y4=(2y)1/4 or (2y)0.25

Caution 6.2.14.

In part (b) of the Example above, the parentheses around (2y) must not be omitted. The expression 2y1/4 means 2y4.
Remember that a negative exponent denotes a reciprocal.

Checkpoint 6.2.15. Practice 5.

Convert each radical to exponential notation.
  1. 1ab5
  2. 3w6
Solution.
  1. (ab)1/5
  2. 3w1/6

Example 6.2.16.

Convert each power to radical notation.
  1. 51/2=5
  2. x0.2=x5
  3. 2x1/3=2x3
  4. 8a1/4=8a4

Note 6.2.17.

In Example 6.2.16, note that the exponent 1/4 applies only to a, not to 8a. Compare with (8a)1/4=18a4  [TK]  

Checkpoint 6.2.18. Practice 6.

  1. Convert 32x4 to exponential notation.
  2. Convert 5b0.125 to radical notation.
Solution.
  1. 3(2x)1/4
  2. 5b8

Checkpoint 6.2.19. QuickCheck 3.

True or false.
  1. All expressions of the form x1/n represent irrational numbers.
  2. When we enter an exponent 1/n into a calculator, we must enclose it in parentheses.
  3. In the expression 16x1/4, the exponent does not apply to 16.
  4. You can use your calculator to find a decimal equivalent for an irrational number.

Subsection 6.2.5 Using Fractional Exponents to Solve Equations

We know that raising to powers and taking roots are inverse operations, that is, each operation undoes the effects of the other. This relationship is especially easy to see when the root is denoted by a fractional exponent. For example, to solve the equation
x5=250
we would take the fifth root of each side. But instead of using radical notation, we can raise both sides of the equation to the power 15:
(x5)1/5=2501/5x3.017
Notice again that (xa)b=xab, so
(x5)1/5=x(1/5)(5)=x1
Thus, to solve an equation involving a power function xn, we first isolate the power, then raise both sides to the exponent 1n.

Example 6.2.20.

For astronomers, the mass of a star is its most important property, but it is also the most difficult to measure directly. For many stars, their luminosity, or brightness, varies roughly as the fourth power of the mass.
  1. Our Sun has luminosity 4×1026 watts and mass 2×1030 kilograms. Because the numbers involved are so large, astronomers often use these solar constants as units of measure: The luminosity of the Sun is 1 solar luminosity, and its mass is 1 solar mass. Write a power function for the luminosity, L, of a star in terms of its mass, M, using units of solar mass and solar luminosity.
  2. The star Sirius is 23 times brighter than the Sun, so its luminosity is 23 solar luminosities. Estimate the mass of Sirius in units of solar mass.

Solution.

  1. Because L varies as the fourth power of M, we have
    L=kM4
    Substituting the values of L and M for the Sun (namely, L=1 and M=1), we find
    1=k(1)4
    so k=1 and L=M4.
  2. We substitute the luminosity of Sirius, L=23, to get
    23=M4
    To solve the equation for M, we raise both sides to the 14 power.
    (23)1/4=(M4)1/42.1899=M
    The mass of Sirius is about 2.2 solar masses, or about 2.2 times the mass of the Sun.

Checkpoint 6.2.21. QuickCheck 4.

  1. How do we solve the equation x8=32?
  2. Which law of exponents justifies this strategy?

Checkpoint 6.2.22. Practice 7.

The manager of the Atlantis Hotel wants to install in the lobby a spherical fish tank that holds at least 900 cubic feet of water. What should be the radius of the fish tank?
Solution.
6 feet

Subsection 6.2.6 Power Functions

The size of a diamond is usually given by its weight w in carats. The diameter, D, of a diamond cut in a traditional round style is then a power function of its weight, given by
D=f(w)=kw1/3

Example 6.2.23.

  1. A diamond weighing one-quarter carat has diameter about 4.05 millimeters. Find the constant of proportionality, k, and write a formula for D as a function of w.
  2. Complete the table with the diameters of diamonds of various weights.
    w  (carats) 0 0.5 0.75 1.0 1.25 1.5 2.0 2.5 3.0
    D  (mm)                                    
  3. Sketch a graph of the function D=f(w).
  4. What is the diameter of the Hope Diamond, which weighs 45.52 carats and is worth $300 million?

Solution.

  1. We substitute w=0.25 and D=4.05 into the equation and solve for k.
    4.05=k(0.25)1/3k=4.050.251/3=6.429
    Rounded to tenths, k=6.4, and D=6.4w1/3.
  2. We evaluate the function for each of the weights given in the table. The diameters are rounded to tenths of a millimeter.
    w  (carats) 0 0.5 0.75 1.0 1.25 1.5 2.0 2.5 3.0
    D  (mm) 0 5.1 5.8 6.4 6.9 7.3 8.1 8.7 9.2
  3. We plot the points from the table and draw a smooth curve to obtain the graph shown.
    diameter vs carats
  4. We evaluate the function for w=45.52 to obtain
    D=6.4(45.52)1/3=22.85
    According to our formula, the Hope Diamond is about 22.85 millimeters in diameter.
Of course, power functions can be expressed using any of the notations we have discussed. For example, the function in the Example above can be written as
f(w)=6.4w1/3    or    f(w)=6.4w0.33    or   f(w)=6.4w3

Checkpoint 6.2.24. Practice 8.

  1. Complete the table of values for the power function f(x)=x1/2.
    x 0.1 0.25 0.5 1 2
    f(x) 000 000 000 000 000
    x 4 8 10 20 200
    f(x) 000 000 000 000 000
  2. Sketch the graph of y=f(x).
  3. Write the formula for f(x) with a decimal exponent, and with radical notation.
Solution.
  1. x 0.1 0.25 0.5 1 2
    f(x) 3.2 2 1.4 1 0.71
    x 4 8 10 20 200
    f(x) 0.5 0.35 0.32 0.22 0.1
  2. power function
  3. f(x)=x0.5, f(x)=1x

Subsection 6.2.7 Solving Radical Equations

A radical equation is one in which the variable appears under a square root or other radical. The radical may be denoted by a fractional exponent. For example, the equation
5x1/3=32
is a radical equation because x1/3=x3. To solve the equation, we first isolate the power to get
x1/3=6.4
Then we raise both sides of the equation to the reciprocal of 13, or 3.
(x1/3)3=6.43x=262.144

Example 6.2.25.

When a car brakes suddenly, its speed can be estimated from the length of the skid marks it leaves on the pavement. A formula for the car’s speed, in miles per hour, is
v=f(d)=(24d)1/2
where the length of the skid marks, d, is given in feet.
  1. If a car leaves skid marks 80 feet long, how fast was the car traveling when the driver applied the brakes?
  2. How far will a car skid if its driver applies the brakes while traveling 80 miles per hour?

Solution.

  1. To find the velocity of the car, we evaluate the function for d=80.
    v=(2480)1/2=(1920)1/243.8178046
    The car was traveling at approximately 44 miles per hour.
  2. We would like to find the value of d when the value of v is known. We substitute v=80 into the formula and solve the equation
    80=(24d)1/2   Solve for d.
    Because d appears to the power 12, we first square both sides of the equation to get
    802=((24d)1/2)2Square both sides.6400=24dDivide by 24.266.6=d
    You can check that this value for d works in the original equation. Thus, the car will skid approximately 267 feet. A graph of the function v=(24d)1/2 is shown below, along with the points corresponding to the values in parts (a) and (b).
    velocity vs braking distance

Note 6.2.26.

Thus, we can solve an equation where one side is an nth root of x by raising both sides of the equation to the nth power. We must be careful when raising both sides of an equation to an even power, since extraneous solutions may be introduced. However, because most applications of power functions deal with positive numbers only, they do not usually involve extraneous solutions.

Checkpoint 6.2.27. QuickCheck 5.

  1. How do we solve the equation  x1/6=12?
  2. How do we solve the equation  x6=12?
  3. How do we solve the equation  x6=12?
  4. How do we evaluate the function  f(x)=x1/6 for x=12?

Checkpoint 6.2.28. Practice 9.

Solve  5(x1)1/4=10
Hint: Isolate the radical, then raise both sides to the fourth power.
Solution.
17

Subsection 6.2.8 Roots of Negative Numbers

You already know that 9 is not a real number, because there is no real number whose square is 9. Similarly, 164 is not a real number, because there is no real number r for which r4=16. (Both of these radicals represent complex numbers.) In general, we cannot find an even root (square root, fourth root, and so on) of a negative number.
On the other hand, every positive number has two even roots that are real numbers. For example, both 3 and 3 are square roots of 9. The symbol 9 refers only to the positive, or principal root, of 9. If we want to refer to the negative square root of 9, we must write 9=3. Similarly, both 2 and 2 are fourth roots of 16, because 24=16 and (2)4=16. However, the symbol 164 refers to the principal, or positive, fourth root only. Thus,
164=2   and   164=2
Things are simpler for odd roots (cube roots, fifth roots, and so on). Every real number, whether positive, negative, or zero, has exactly one real-valued odd root. For example,
325=2   and   325=2
Here is a summary of our discussion.

Roots of Real Numbers.

  1. Every positive number has two real-valued roots, one positive and one negative, if the index is even.
  2. A negative number has no real-valued root if the index is even.
  3. Every real number, positive, negative, or zero, has exactly one real-valued root if the index is odd.

Example 6.2.29.

  1. 6254 is not a real number.
  2. 6254=5
  3. 15=1
  4. 14 is not a real number.
The same principles apply to powers with fractional exponents. Thus
(32)1/5=2
but (64)1/6 is not a real number. On the other hand,
641/6=2
because the exponent 1/6 applies only to 64, and the negative sign is applied after the root is computed.

Checkpoint 6.2.30. Practice 10.

Evaluate each power, if possible.
  1. 811/4
  2. (81)1/4
  3. 641/3
  4. (64)1/3
Solution.
  1. 3
  2. undefined
  3. 4
  4. 4

Exercises 6.2.9 Problem Set 6.2

Warm Up

1.
Each radical below is equal to an integer. Use trial and error to find the root without a calculator.
  1. 169
  2. 643
  3. 814
  4. 100,0005
  5. 12964
  6. 3433
2.
Evaluate each power without using a calculator.
  1. 91/2
  2. 811/4
  3. 641/6
  4. 320.2
  5. 81/3
  6. 640.5
3.
Use a calculator to approximate each irrational number to the nearest thousandth:
  1. 21/2
  2. 753
  3. 1.64
  4. 3651/3
  5. 0.0060.2
  6. 1000.25
4.
Use the definition of a root to simplify without using a calculator.
  1. (1253)3=
  2. (24)4=
  3. (73)3=
  4. (2123)3
  5. (a3a24)4
  6. (x22x3)3=

Skills Practice

5.
Evaluate each power, if possble.
  1. 811/4
  2. (81)1/4
  3. 271/3
  4. (27)1/3
6.
Complete the table converting radicals to powers.
Radical x x3 x4 x5 1x 1x3 1x4 1x5
Exponent (Fraction) x1/2                            
Exponent (Decimal) x0.5                            
Exercise Group.
For Problems 7 and 8, write each expression in radical form.
8.
  1. 81/4
  2. y(5x)0.5
  3. (y+2)1/3
Exercise Group.
For Problems 9 and 10, write each expression in exponential form.
Exercise Group.
For Problems 11–14, write the expression as a sum of terms of the form kxp.
14.
12x1+2x0.5+x0.252x0.5
Exercise Group.
For Problems 15–20, solve.
Exercise Group.
For Problems 21–26, solve for the indicated variable.
21.
T=2πLg     for L
22.
r=t2s2     for s
23.
d=16Mr2m3     for M
24.
R=8Lvfπp4     for p
25.
T=ESA4     for A
26.
r=3V4π3     for V
27.
Match each function with the description of its graph in the first quadrant.
  1. f(x)=x2
  2. g(x)=x2
  3. h(x)=x1/2
  4. f(x)=x1/2
  1. Increasing and concave up
  2. Increasing and concave down
  3. Decreasing and concave up
  4. Decreasing and concave down
28.
Match each function with its graph.
  1. cube and square
    f(x)=x2, g(x)=x3
  2. cube root and square root
    f(x)=x1/2, g(x)=x1/3
  3. inverse cube and inverse square
    f(x)=x2, g(x)=x3
29.
Write each expression as a power function.
  1. G(x)=3.7x3
  2. H(x)=85x4
  3. F(t)=25t5
30.
Graph each set of functions in the given window. What do you observe?
  1. y1=x, y2=x2, y3=x
    Xmin=0Xmax=4Ymin=0Ymax=4
  2. y1=x3, y2=x3, y3=x
    Xmin=4Xmax=4Ymin=4Ymax=4

Applications

31.
When a ship moves through the water, it creates waves that hinder its own progress. Because of this resistance, there is an upper limit to the speed at which a ship can travel, given in knots by
vmax=1.3L
where L is the length of the vessel in feet. (Source: Gilner, 1972)
  1. Complete the table of values for vmax as a function of L.
    L (feet) 200 400 600 800 1000
    vmax (knots)
  2. Graph maximum speed against vessel length.
    grid
  3. The largest ship ever built, the oil tanker Seawise Giant, was 1,504 feet long. What was its top speed?
  4. As a ship approaches its maximum speed, the power required increases sharply. Therefore, most merchant ships are designed to cruise at speeds no higher than
    vc=0.8L
    Graph vc on the same axes with vmax.
  5. What was the cruising speed of the Seawise Giant? What percent of its maximum speed is that?
32.
If you walk in the normal way, your speed, v, in meters per second, is limited by the length of your legs, r, according to the formula
vgr
where g is the acceleration due to gravity. (Source: Alexander, 1992)
  1. Complete the table of values for v as a function of r. The constant, g, is approximately 10.
    r (meters) 0.2 0.4 0.6 0.8 1.0
    v (meters/sec)
  2. A typical adult man has legs about 0.9 meter long. How fast can he walk?
  3. A typical four-year-old has legs 0.5 meter long. How fast can she walk?
  4. Graph maximum walking speed against leg length.
    grid
  5. Race-walkers can walk as fast as 4.4 meters per second by rotating their hips so that the effective length of their legs is increased. What is that effective length?
  6. On the moon the value of g is 1.6. How fast can a typical adult man walk on the moon?
33.
The period of a pendulum is the time it takes for the pendulum to complete one entire swing, from left to right and back again. The greater the length, L, of the pendulum, the longer its period, T. In fact, if L is measured in feet, then the period is given in seconds by
T=2πL32
  1. Write the formula for T as a power function in the form f(x)=kxp.
  2. The Foucault pendulum in the Convention Center in Portland, Oregon is 90 feet long. What is its period?
  3. Choose a suitable window and graph the function T=f(L). Label the point corresponding to point (b) on the graph.
34.
The rate, r, in feet per second, at which water flows from a fire hose varies directly with the square root of the water pressure, P, in psi (pounds per square inch). What is the rate of water flow at a typical water pressure of 60 psi?
P (psi) 10 20 30 40
r (ft/sec) 38.3 54.1 66.3 76.5
  1. Find the value of k and write a power function relating the variables.
  2. What is the rate of water flow at a typical water pressure of 60 psi?
  3. Graph your function and verify your answer to part (b) on the graph.
35.
Thanks to improvements in technology, the annual electricity cost of running most major appliances has decreased steadily since 1970. The average annual cost of running a refrigerator is given, in dollars, by the function
C(t)=14828t1/3
where t is the number of years since 1970.
  1. How much did it cost to run a refrigerator in 1980? In 1990?
  2. When was the cost of running a refrigerator half of the cost in 1970? If the cost continues to decline according to the given function, when will it cost $50 per year to run a refrigerator?
  3. Graph the function C(t). Do you think that the cost will continue to decline indefinitely according to the given function? Why or why not?
36.
A rough estimate for the radius of the nucleus of an atom is provided by the formula
r=kA1/3
where A is the mass number of the nucleus and k1.3×1013 centimeter.
  1. Estimate the radius of the nucleus of an atom of iodine-127, which has mass number 127. If the nucleus is roughly spherical, what is its volume?
  2. The nuclear mass of iodine-127 is 2.1×1022 gram. What is the density of the nucleus? (Density is mass per unit volume.)
  3. Complete the table of values for the radii of various radioisotopes.
    Element Carbon Potassium Cobalt Technetium Radium
    Mass
    number, A
    14 40 60 99 226
    Radius, r
  4. Sketch a graph of r as a function of A. (Use units of 1013 centimeter on the vertical axis.)
37.
When the Concorde lands at Heathrow airport in London, the width w of the sonic boom felt on the ground is given in kilometers by the following formula:
w=4(Thm)1/2
where T stands for the temperature on the ground in kelvins, h is the altitude of the Concorde when it breaks the sound barrier, and m is the drop in temperature for each gain in altitude of 1 kilometer. Find the width of the sonic boom if the ground temperature is 293 kelvins, the altitude of the Concorde is 15 kilometers, and the temperature drop is 4 kelvin per kilometer of altitude.
38.
The Stefan-Boltzmann law says that the temperature, T, of the Sun, in kelvins, can be computed from the formula
sT4=L4πR2
where L=3.9×1033 is the total luminosity of the Sun, R=9.96×1010 centimeters is the radius of the Sun, and s=5.7×105 is a constant governing radiation. Calculate the temperature of the Sun.
39.
  1. Graph the functions   f(x)=4x93   and   g(x)=12   in the window
    Xmin=0Xmax=47Ymin=8Ymax=16
  2. Use the graph to solve the equation 4x93=12.
  3. Solve the equation algebraically.
40.
  1. Graph the functions   f(x)=6+212x4   and   g(x)=10   in the window
    Xmin=27Xmax=20Ymin=4Ymax=12
  2. Use the graph to solve the equation 6+212x4=10.
  3. Solve the equation algebraically.
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