College Algebra

Section 0.9: Radicals and Equations from Precalculus Prerequisites a.k.a. ‘Chapter 0’ by Carl Stitz, PhD, and Jeff Zeager, PhD, is available under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 license. © 2013, Carl Stitz. UMGC has modified this work and it is available under the original license.

0.9 Radicals and Equations 111

0.9 Radicals and Equations

In this section we review simplifying expressions and solving equations involving radicals. In addi- tion to the product, quotient and power rules stated in Theorem 0.1 in Section 0.2, we present the following result which states that nth roots and nth powers more or less ‘undo’ each other.1

Theorem 0.11. Simplifying nth powers of nth roots: Suppose n is a natural number, a is a real number and n

p a is a real number. Then

• ( n p

a)n = a

• if n is odd, n p

an = a; if n is even, n p

an = |a|.

Since n p

a is defined so that ( n p

a)n = a, the first claim in the theorem is just a re-wording of Definition 0.8. The second part of the theorem breaks down along odd/even exponent lines due to how exponents affect negatives. To see this, consider the specific cases of 3

p

(�2)3 and 4 p

(�2)4.

In the first case, 3 p

(�2)3 = 3 p �8 = �2, so we have an instance of when n

p an = a. The reason that

the cube root ‘undoes’ the third power in 3 p

(�2)3 = �2 is because the negative is preserved when raised to the third (odd) power. In 4

p

(�2)4, the negative ‘goes away’ when raised to the fourth (even) power: 4

p

(�2)4 = 4 p

16. According to Definition 0.8, the fourth root is defined to give only non-negative numbers, so 4

p 16 = 2. Here we have a case where 4

p

(�2)4 = 2 = | � 2|, not �2.

In general, we need the absolute values to simplify n p

an only when n is even because a negative to an even power is always positive. In particular,

p x 2 = |x|, not just ‘x ’ (unless we know x � 0.)2

We practice these formulas in the following example.

Example 0.9.1. Perform the indicated operations and simplify.

1. p

x 2 + 1 2. p

t 2 � 10t + 25 3. 3 p

48x 14 4. 4

r

⇡r 4

L8

5. 2x 3 p

x 2 � 4 + 2 ✓

1 2( 3 p

x 2 � 4)2

◆

(2x ) 6. q

( p

18y � p

8y )2 + ( p

20 � p

80)2

Solution.

1. We told you back on page 32 that roots do not ‘distribute’ across addition and since x 2 + 1 cannot be factored over the real numbers,

p x 2 + 1 cannot be simplified. It may seem silly

to start with this example but it is extremely important that you understand what maneuvers are legal and which ones are not.3

1See Section 5.3 for a more precise understanding of what we mean here. 2If this discussion sounds familiar, see the discussion following Definition 0.9 and the discussion following ‘Extracting

the Square Root’ on page 83. 3You really do need to understand this otherwise horrible evil will plague your future studies in Math. If you say

something totally wrong like p

x 2 + 1 = x + 1 then you may never pass Calculus. PLEASE be careful!

112 Prerequisites

2. Again we note that p

t 2 � 10t + 25 6= p

t 2� p

10t + p

25, since radicals do not distribute across addition and subtraction.4 In this case, however, we can factor the radicand and simplify as

p

t 2 � 10t + 25 = p

(t � 5)2 = |t � 5|

Without knowing more about the value of t , we have no idea if t � 5 is positive or negative so |t � 5| is our final answer.5

3. To simplify 3 p

48x 14, we need to look for perfect cubes in the radicand. For the cofficient, we have 48 = 8 · 6 = 23 · 6. To find the largest perfect cube factor in x 14, we divide 14 (the exponent on x ) by 3 (since we are looking for a perfect cube). We get 4 with a remainder of 2. This means 14 = 4 · 3 + 2, so x 14 = x 4·3+2 = x 4·3x 2 = (x 4)3x 2. Putting this altogether gives:

3 p

48x 14 = 3 p

23 · 6 · (x 4)3x 2 Factor out perfect cubes = 3

p 23 3 p

(x 4)3 3 p

6x 2 Rearrange factors, Product Rule of Radicals = 2x 4 3

p 6x 2

4. In this example, we are looking for perfect fourth powers in the radicand. In the numerator r 4 is clearly a perfect fourth power. For the denominator, we take the power on the L, namely 12, and divide by 4 to get 3. This means L8 = L2·4 = (L2)4. We get

4

r

⇡r 4

L12 =

4 p

⇡r 4 4 p

L12 Quotient Rule of Radicals

= 4 p

⇡ 4 p

r 4 4 p

(L2)4 Product Rule of Radicals

= 4 p

⇡|r| |L2|

Simplify

Without more information about r , we cannot simplify |r| any further. However, we can simplify |L2|. Regardless of the choice of L, L2 � 0. Actually, L2 > 0 because L is in the denominator which means L 6= 0. Hence, |L2| = L2. Our answer simplifies to:

4 p

⇡|r| |L2|

= |r| 4

p ⇡

L2

5. After a quick cancellation (two of the 2’s in the second term) we need to obtain a common denominator. Since we can view the first term as having a denominator of 1, the common denominator is precisely the denominator of the second term, namely ( 3

p x 2 � 4)2. With

4Let t = 1 and see what happens to p

t 2 � 10t + 25 versus p

t 2 � p

10t + p

25. 5In general, |t � 5| 6= |t| � |5| and |t � 5| 6= t + 5 so watch what you’re doing!

0.9 Radicals and Equations 113

common denominators, we proceed to add the two fractions. Our last step is to factor the numerator to see if there are any cancellation opportunities with the denominator.

2x 3 p

x 2 � 4 + 2 ✓

1 2( 3 p

x 2 � 4)2

◆

(2x ) = 2x 3 p

x 2 � 4 + ◆2 ✓

1 ◆2(

3 p

x 2 � 4)2

◆

(2x ) Reduce

= 2x 3 p

x 2 � 4 + 2x

( 3 p

x 2 � 4)2 Mutiply

= (2x 3 p

x 2 � 4) · ( 3 p

x 2 � 4)2

( 3 p

x 2 � 4)2 +

2x ( 3 p

x 2 � 4)2 Equivalent

fractions

= 2x ( 3

p x 2 � 4)3

( 3 p

x 2 � 4)2 +

2x ( 3 p

x 2 � 4)2 Multiply

= 2x (x 2 � 4) ( 3 p

x 2 � 4)2 +

2x ( 3 p

x 2 � 4)2 Simplify

= 2x (x 2 � 4) + 2x

( 3 p

x 2 � 4)2 Add

= 2x (x 2 � 4 + 1)

( 3 p

x 2 � 4)2 Factor

= 2x (x 2 � 3) ( 3 p

x 2 � 4)2

We cannot reduce this any further because x 2 � 3 is irreducible over the rational numbers.

6. We begin by working inside each set of parentheses, using the product rule for radicals and combining like terms. q

( p

18y � p

8y )2 + ( p

20 � p

80)2 = q

( p

9 · 2y � p

4 · 2y )2 + ( p

4 · 5 � p

16 · 5)2

= q

( p

9 p

2y � p

4 p

2y )2 + ( p

4 p

5 � p

16 p

5)2

= q

(3 p

2y � 2 p

2y )2 + (2 p

5 � 4 p

5)2

= q

( p

2y )2 + (�2 p

5)2

= q

2y + (�2)2( p

5)2

= p

2y + 4 · 5

= p

2y + 20

To see if this simplifies any further, we factor the radicand: p

2y + 20 = p

2(y + 10). Finding no perfect square factors, we are done.

114 Prerequisites

Theorem 0.11 allows us to generalize the process of ‘Extracting Square Roots’ to ‘Extracting nth roots’ which in turn allows us to solve equations6 of the form X n = c.

Extracting nth roots:

• If c is a real number and n is odd then the real number solution to X n = c is X = n p

c.

• If c � 0 and n is even then the real number solutions to X n = c are X = ± n p

c.

Note: If c < 0 and n is even then X n = c has no real number solutions.

Essentially, we solve X n = c by ‘taking the nth root’ of both sides: n p

X n = n p

c. Simplifying the left side gives us just X if n is odd or |X | if n is even. In the first case, X = n

p c, and in the second,

X = ± n p

c. Putting this together with the other part of Theorem 0.11, namely ( n p

a)n = a, gives us a strategy for solving equations which involve nth and nth roots.

Strategies for Power and Radical Equations

• If the equation involves an nth power and the variable appears in only one term, isolate the term with the nth power and extract nth roots.

• If the equation involves an nth root and the variable appears in that nth root, isolate the nth root and raise both sides of the equation to the nth power.

Note: When raising both sides of an equation to an even power, be sure to check for extraneous solutions.

The note about ‘extraneous solutions’ can be demonstrated by the basic equation: p

x = �2. This equation has no solution since, by definition,

p x � 0 for all real numbers x . However, if we square

both sides of this equation, we get ( p

x )2 = (�2)2 or x = 4. However, x = 4 doesn’t check in the original equation, since

p 4 = 2, not �2. Once again, the root7 of all of our problems lies in the

fact that a negative number to an even power results in a positive number. In other words, raising both sides of an equation to an even power does not produce an equivalent equation, but rather, an equation which may possess more solutions than the original. Hence the cautionary remark above about extraneous solutions.

Example 0.9.2. Solve the following equations.

1. (5x + 3)4 = 16 2. 1 � (5 � 2w )3

7 = 9 3. t +

p 2t + 3 = 6

4. p

2 � 3 3 p

2y + 1 = 0 5. p

4x � 1 + 2 p

1 � 2x = 1 6. 4 p

n2 + 2 + n = 0

For the remaining problems, assume that all of the variables represent positive real numbers.8

6Well, not entirely. The equation x 7 = 1 has seven answers: x = 1 and six complex number solutions which we’ll find using techniques in Section 11.7.

7Pun intended! 8That is, you needn’t worry that you’re multiplying or dividing by 0 or that you’re forgetting absolute value symbols.

0.9 Radicals and Equations 115

7. Solve for r : V = 4⇡ 3

(R3 � r 3). 8. Solve for M1: r1 r2

= r

M2 M1

9. Solve for v : m = m0

r

1 � v 2

c2

. Assume all quantities represent positive real numbers.

Solution.

1. In our first equation, the quantity containing x is already isolated, so we extract fourth roots. Since the exponent here is even, when the roots are extracted we need both the positive and negative roots.

(5x + 3)4 = 16 5x + 3 = ± 4

p 16 Extract fourth roots

5x + 3 = ±2 5x + 3 = 2 or 5x + 3 = �2

x = � 1 5

or x = �1

We leave it to the reader that both of these solutions satisfy the original equation.

2. In this example, we first need to isolate the quantity containing the variable w . Here, third (cube) roots are required and since the exponent (index) is odd, we do not need the ±:

1 � (5 � 2w )3

7 = 9

� (5 � 2w )3

7 = 8 Subtract 1

(5 � 2w )3 = �56 Multiply by �7 5 � 2w = 3

p �56 Extract cube root

5 � 2w = 3 p

(�8)(7) 5 � 2w = 3

p �8 3

p 7 Product Rule

5 � 2w = �2 3 p

7 �2w = �5 � 2 3

p 7 Subtract 5

w = �5 � 2 3

p 7

�2 Divide by �2

w = 5 + 2 3

p 7

2 Properties of Negatives

The reader should check the answer because it provides a hearty review of arithmetic.

3. To solve t + p

2t + 3 = 6, we first isolate the square root, then proceed to square both sides of the equation. In doing so, we run the risk of introducing extraneous solutions so checking

116 Prerequisites

our answers here is a necessity.

t + p

2t + 3 = 6 p

2t + 3 = 6 � t Subtract t ( p

2t + 3)2 = (6 � t )2 Square both sides 2t + 3 = 36 � 12t + t 2 F.O.I.L. / Perfect Square Trinomial

0 = t 2 � 14t + 33 Subtract 2t and 3 0 = (t � 3)(t � 11) Factor

From the Zero Product Property, we know either t � 3 = 0 (which gives t = 3) or t � 11 = 0 (which gives t = 11). When checking our answers, we find t = 3 satisfies the original equation, but t = 11 does not.9 So our final answer is t = 3 only.

4. In our next example, we locate the variable (in this case y ) beneath a cube root, so we first isolate that root and cube both sides.

p 2 � 3 3

p

2y + 1 = 0 �3 3 p

2y + 1 = � p

2 Subtract p

2

3 p

2y + 1 = � p

2 �3

Divide by �3

3 p

2y + 1 = p

2 3

Properties of Negatives

( 3 p

2y + 1)3 =

p 2

3

!3

Cube both sides

2y + 1 = ( p

2)3

33

2y + 1 = 2 p

2 27

2y = 2 p

2 27

� 1 Subtract 1

2y = 2 p

2 27

� 27 27

Common denominators

2y = 2 p

2 � 27 27

Subtract fractions

y = 2 p

2 � 27 54

Divide by 2 �

multiply by 12 �

Since we raised both sides to an odd power, we don’t need to worry about extraneous solutions but we encourage the reader to check the solution just for the fun of it.

9It is worth noting that when t = 11 is substituted into the original equation, we get 11 + p

25 = 6. If the + p

25 were � p

25, the solution would check. Once again, when squaring both sides of an equation, we lose track of ±, which is what lets extraneous solutions in the door.

0.9 Radicals and Equations 117

5. In the equation p

4x � 1 + 2 p

1 � 2x = 1, we have not one but two square roots. We begin by isolating one of the square roots and squaring both sides. p

4x � 1 + 2 p

1 � 2x = 1 p

4x � 1 = 1 � 2 p

1 � 2x Subtract 2 p

1 � 2x from both sides ( p

4x � 1)2 = (1 � 2 p

1 � 2x )2 Square both sides 4x � 1 = 1 � 4

p 1 � 2x + (2

p 1 � 2x )2 F.O.I.L. / Perfect Square Trinomial

4x � 1 = 1 � 4 p

1 � 2x + 4(1 � 2x ) 4x � 1 = 1 � 4

p 1 � 2x + 4 � 8x Distribute

4x � 1 = 5 � 8x � 4 p

1 � 2x Gather like terms

At this point, we have just one square root so we proceed to isolate it and square both sides a second time.10

4x � 1 = 5 � 8x � 4 p

1 � 2x 12x � 6 = �4

p 1 � 2x Subtract 5, add 8x

(12x � 6)2 = (�4 p

1 � 2x )2 Square both sides 144x 2 � 144x + 36 = 16(1 � 2x ) 144x 2 � 144x + 36 = 16 � 32x 144x 2 � 112x + 20 = 0 Subtract 16, add 32x 4(36x 2 � 28x + 5) = 0 Factor

4(2x � 1)(18x � 5) = 0 Factor some more

From the Zero Product Property, we know either 2x � 1 = 0 or 18x � 5 = 0. The former gives x = 12 while the latter gives us x =

5 18 . Since we squared both sides of the equation (twice!),

we need to check for extraneous solutions. We find x = 518 to be extraneous, so our only solution is x = 12 .

6. As usual, our first step in solving 4 p

n2 + 2 + n = 0 is to isolate the radical. We then proceed to raise both sides to the fourth power to eliminate the fourth root:

4 p

n2 + 2 + n = 0 4 p

n2 + 2 = �n Subtract n ( 4 p

n2 + 2)4 = (�n)4 Raise both sides to the 4th power n2 + 2 = n4 Properties of Negatives

0 = n4 � n2 � 2 Subtract n2 and 2 0 = (n2 � 2)(n2 + 1) Factor - this is a ‘Quadratic in Disguise’

At this point, the Zero Product Property gives either n2 � 2 = 0 or n2 + 1 = 0. From n2 � 2 = 0, we get n2 = 2, so n = ±

p 2. From n2 + 1 = 0, we get n2 = �1, which gives no real solutions.11

10To avoid complications with fractions, we’ll forego dividing by the coefficient of p

1 � 2x , namely �4. This is perfectly fine so long as we don’t forget to square it when we square both sides of the equation.

11Why is that again?

118 Prerequisites

Since we raised both sides to an even (the fourth) power, we need to check for extraneous solutions. We find that n = �

p 2 works but n =

p 2 is extraneous.

7. In this problem, we are asked to solve for r . While there are a lot of letters in this equation12, r appears in only one term: r 3. Our strategy is to isolate r 3 then extract the cube root.

V = 4⇡ 3

(R3 � r 3) 3V = 4⇡(R3 � r 3) Multiply by 3 to clear fractions 3V = 4⇡R3 � 4⇡r 3 Distribute

3V � 4⇡R3 = �4⇡r 3 Subtract 4⇡R3

3V � 4⇡R3

�4⇡ = r 3 Divide by �4⇡

4⇡R3 � 3V 4⇡

= r 3 Properties of Negatives

3

r

4⇡R3 � 3V 4⇡

= r Extract the cube root

The check is, as always, left to the reader and highly encouraged.

8. The equation we are asked to solve in this example is from the world of Chemistry and is none other than Graham’s Law of effusion. As was mentioned in Example 0.8.2, subscripts in Mathematics are used to distinguish between variables and have no arithmetic signifi- cance. In this example, r1, r2, M1 and M2 are as different as x , y , z and 117. Since we are asked to solve for M1, we locate M1 and see it is in a denominator in a square root. We eliminate the square root by squaring both sides and proceed from there.

r1 r2

= r

M2 M1

✓

r1 r2

◆2 = ✓

r

M2 M1

◆2

Square both sides

r 21 r 22

= M2 M1

r 21 M1 = M2r 2 2 Multiply by r

2 2 M1 to clear fractions, assume r2, M1 6= 0

M1 = M2r 22

r 21 Divide by r 21 , assume r1 6= 0

As the reader may expect, checking the answer amounts to a good exercise in simplifying rational and radical expressions. The fact that we are assuming all of the variables represent positive real numbers comes in to play, as well.

12including a Greek letter, no less!

0.9 Radicals and Equations 119

9. Our last equation to solve comes from Einstein’s Special Theory of Relativity and relates the mass of an object to its velocity as it moves.13 We are asked to solve for v which is located in just one term, namely v 2, which happens to lie in a fraction underneath a square root which is itself a denominator. We have quite a lot of work ahead of us!

m = m0

r

1 � v 2

c2

m

r

1 � v 2

c2 = m0 Multiply by

r

1 � v 2

c2 to clear fractions

m

r

1 � v 2

c2

!2

= m20 Square both sides

m2 ✓

1 � v 2

c2

◆

= m20 Properties of Exponents

m2 � m2v 2

c2 = m20 Distribute

� m2v 2

c2 = m20 � m2 Subtract m2

m2v 2 = �c2(m20 � m2) Multiply by �c2 (c2 6= 0) m2v 2 = �c2m20 + c2m2 Distribute

v 2 = c2m2 � c2m20

m2 Rearrange terms, divide by m2 (m2 6= 0)

v =

r

c2m2 � c2m20 m2

Extract Square Roots, v > 0 so no ±

v = p

c2(m2 � m20 )p m2

Properties of Radicals, factor

v = |c| p

m2 � m20 |m|

v = c p

m2 � m20 m

c > 0 and m > 0 so |c| = c and |m| = m

Checking the answer algebraically would earn the reader great honor and respect on the Algebra battlefield so it is highly recommended.

0.9.1 Rationalizing Denominators and Numerators

In Section 0.7, there were a few instances where we needed to ‘rationalize’ a denominator - that is, take a fraction with radical in the denominator and re-write it as an equivalent fraction without

13See this article on the Lorentz Factor.

120 Prerequisites

a radical in the denominator. There are various reasons for wanting to do this,14 but the most pressing reason is that rationalizing denominators - and numerators as well - gives us an oppor- tunity for more practice with fractions and radicals. To help refresh your memory, we rationalize a denominator and then a numerator below:

1 p

2 =

p 2

p 2 p

2 = p

2 p

4 = p

2 2

and 7 3 p

4 3

= 7 3 p

4 3 p

2 3 3 p

2 =

7 3 p

8 3 3 p

2 =

7 · 2 3 3 p

2 =

14 3 3 p

2

In general, if the fraction contains either a single term numerator or denominator with an unde- sirable nth root, we multiply the numerator and denominator by whatever is required to obtain a perfect nth power in the radicand that we want to eliminate. If the fraction contains two terms the situation is somewhat more complicated. To see why, consider the fraction 3

4� p

5 . Suppose we

wanted to rid the denominator of the p

5 term. We could try as above and multiply numerator and denominator by

p 5 but that just yields:

3 4 �

p 5

= 3 p

5 (4 �

p 5) p

5 =

3 p

5 4 p

5 � p

5 p

5 =

3 p

5 4 p

5 � 5

We haven’t removed p

5 from the denominator - we’ve just shuffled it over to the other term in the denominator. As you may recall, the strategy here is to multiply both numerator and denominator by what’s called the conjugate.

Definition 0.17. Congugate of a Square Root Expression: If a, b and c are real numbers with c > 0 then the quantities (a + b

p c) and (a � b

p c) are conjugates of one another.a Conjugates

multiply according to the Difference of Squares Formula:

(a + b p

c)(a � b p

c) = a2 � (b p

c)2 = a2 � b2c aAs are (b

p c � a) and (b

p c + a): (b

p c � a)(b

p c + a) = b2c � a2.

That is, to get the conjugate of a two-term expression involving a square root, you change the ‘�’ to a ‘+,’ or vice-versa. For example, the conjugate of 4 �

p 5 is 4 +

p 5, and when we multiply these

two factors together, we get (4 � p

5)(4 + p

5) = 42 � ( p

5)2 = 16 � 5 = 11. Hence, to eliminate thep 5 from the denominator of our original fraction, we multiply both the numerator and denominator

by the conjugate of 4 � p

5:

3 4 �

p 5

= 3(4 +

p 5)

(4 � p

5)(4 + p

5) =

3(4 + p

5) 42 � (

p 5)2

= 3(4 +

p 5)

16 � 5 =

12 + 3 p

5 11

What if we had 3 p

5 instead of p

5? We could try multiplying 4 � 3 p

5 by 4 + 3 p

5 to get

(4 � 3 p

5)(4 + 3 p

5) = 42 � ( 3 p

5)2 = 16 � 3 p

25, 14Before the advent of the handheld calculator, rationalizing denominators made it easier to get decimal approxima-

tions to fractions containing radicals. However, some (admittedly more abstract) applications remain today – one of which we’ll explore in Section 0.10; one you’ll see in Calculus.

0.9 Radicals and Equations 121

which leaves us with a cube root. What we need to undo the cube root is a perfect cube, which means we look to the Difference of Cubes Formula for inspiration: a3 � b3 = (a � b)(a2 + ab + b2). If we take a = 4 and b = 3

p 5, we multiply

(4 � 3 p

5)(42 + 4 3 p

5 + ( 3 p

5)2) = 43 + 42 3 p

5 + 4 3 p

5 � 42 3 p

5 � 4( 3 p

5)2 � ( 3 p

5)3 = 64 � 5 = 59

So if we were charged with rationalizing the denominator of 3 4� 3

p 5 , we’d have:

3 4 � 3

p 5

= 3(42 + 4 3

p 5 + ( 3

p 5)2)

(4 � 3 p

5)(42 + 4 3 p

5 + ( 3 p

5)2) =

48 + 12 3 p

5 + 3 3 p

25 59

This sort of thing extends to nth roots since (a � b) is a factor of an � bn for all natural numbers n, but in practice, we’ll stick with square roots with just a few cube roots thrown in for a challenge.15

Example 0.9.3. Rationalize the indicated numerator or denominator:

1. Rationalize the denominator: 3

5 p

24x 2 2. Rationalize the numerator:

p 9 + h � 3

h

Solution.

1. We are asked to rationalize the denominator, which in this case contains a fifth root. That means we need to work to create fifth powers of each of the factors of the radicand. To do so, we first factor the radicand: 24x 2 = 8 · 3 · x 2 = 23 · 3 · x 2. To obtain fifth powers, we need to multiply by 22 · 34 · x 3 inside the radical.

3 6 p

24x 2 =

3 5 p

23 · 3 · x 2

= 3 5 p

22 · 34 · x 3 5 p

23 · 3 · x 2 5 p

22 · 34 · x 3 Equivalent Fractions

= 3 5 p

22 · 34 · x 3 5 p

23 · 3 · x 2 · 22 · 34 · x 3 Product Rule

= 3 5 p

22 · 34 · x 3 5 p

25 · 35 · x 5 Property of Exponents

= 3 5 p

22 · 34 · x 3 5 p

25 5 p

35 5 p

x 5 Product Rule

= 3 5 p

22 · 34 · x 3 2 · 3 · x

Product Rule

= ◆ 3 5 p

4 · 81 · x 3 2 · ◆3 · x

Reduce

= 5 p

324x 3

2x Simplify

15To see what to do about fourth roots, use long division to find (a4 � b4) ÷ (a � b), and apply this to 4 � 4 p

5.

122 Prerequisites

2. Here, we are asked to rationalize the numerator. Since it is a two term numerator involving a square root, we multiply both numerator and denominator by the conjugate of

p 9 + h � 3,

namely p

9 + h + 3. After simplifying, we find an opportunity to reduce the fraction: p

9 + h � 3 h

= ( p

9 + h � 3)( p

9 + h + 3) h( p

9 + h + 3) Equivalent Fractions

= ( p

9 + h)2 � 32

h( p

9 + h + 3) Difference of Squares

= (9 + h) � 9

h( p

9 + h + 3) Simplify

= h

h( p

9 + h + 3) Simplify

= � �� 1

h ◆h( p

9 + h + 3) Reduce

= 1

p 9 + h + 3

We close this section with an awesome example from Calculus.

Example 0.9.4. Simplify the compound fraction

1 p

2(x + h) + 1 �

1 p

2x + 1 h

then rationalize the nu- merator of the result.

Solution. We start by multiplying the top and bottom of the ‘big’ fraction by p

2x + 2h + 1 p

2x + 1.

1 p

2(x + h) + 1 �

1 p

2x + 1 h

=

1 p

2x + 2h + 1 �

1 p

2x + 1 h

=

✓

1 p

2x + 2h + 1 �

1 p

2x + 1

◆p 2x + 2h + 1

p 2x + 1

h p

2x + 2h + 1 p

2x + 1

=

(((( (((p2x + 2h + 1

p 2x + 1

(((( (((p2x + 2h + 1

� p

2x + 2h + 1⇠⇠⇠ ⇠⇠p

2x + 1 ⇠⇠⇠

⇠⇠p 2x + 1

h p

2x + 2h + 1 p

2x + 1

= p

2x + 1 � p

2x + 2h + 1 h p

2x + 2h + 1 p

2x + 1

Next, we multiply the numerator and denominator by the conjugate of p

2x + 1 � p

2x + 2h + 1,

0.9 Radicals and Equations 123

namely p

2x + 1 + p

2x + 2h + 1, simplify and reduce: p

2x + 1 � p

2x + 2h + 1 h p

2x + 2h + 1 p

2x + 1 =

( p

2x + 1 � p

2x + 2h + 1)( p

2x + 1 + p

2x + 2h + 1) h p

2x + 2h + 1 p

2x + 1( p

2x + 1 + p

2x + 2h + 1)

= ( p

2x + 1)2 � ( p

2x + 2h + 1)2

h p

2x + 2h + 1 p

2x + 1( p

2x + 1 + p

2x + 2h + 1)

= (2x + 1) � (2x + 2h + 1)

h p

2x + 2h + 1 p

2x + 1( p

2x + 1 + p

2x + 2h + 1)

= 2x + 1 � 2x � 2h � 1

h p

2x + 2h + 1 p

2x + 1( p

2x + 1 + p

2x + 2h + 1)

= �2◆h

◆h p

2x + 2h + 1 p

2x + 1( p

2x + 1 + p

2x + 2h + 1)

= �2

p 2x + 2h + 1

p 2x + 1(

p 2x + 1 +

p 2x + 2h + 1)

While the denominator is quite a bit more complicated than what we started with, we have done what was asked of us. In the interest of full disclosure, the reason we did all of this was to cancel the original ‘h’ from the denominator. That’s an awful lot of effort to get rid of just one little h, but you’ll see the significance of this in Calculus.

124 Prerequisites

0.9.2 Exercises

In Exercises 1 - 13, perform the indicated operations and simplify.

1. p

9x 2 2. 3 p

8t 3 3. p

50y 6

4. p

4t 2 + 4t + 1 5. p

w 2 � 16w + 64 6. q

( p

12x � p

3x )2 + 1

7.

r

c2 � v 2

c2 8.

3

r

24⇡r 5

L3 9. 4

s

32⇡"8

⇢12

10. p

x � x + 1 p

x 11. 3

p 1 � t 2 + 3t

✓

1 2 p

1 � t 2

◆

(�2t )

12. 2 3 p

1 � z + 2z

1

3 �

3 p

1 � z �2

!

(�1) 13. 3

3 p

2x � 1 + (3x )

� 1

3 �

3 p

2x � 1 �4

!

(2)

In Exercises 14 - 25, find all real solutions.

14. (2x + 1)3 + 8 = 0 15. (1 � 2y )4

3 = 27 16.

1 1 + 2t 3

= 4

17. p

3x + 1 = 4 18. 5 � 3 p

t 2 + 1 = 1 19. x + 1 = p

3x + 7

20. y + p

3y + 10 = �2 21. 3t + p

6 � 9t = 2 22. 2x � 1 = p

x + 3

23. w = 4 p

12 � w 2 24. p

x � 2 + p

x � 5 = 3 25. p

2x + 1 = 3 + p

4 � x

In Exercises 26 - 29, solve each equation for the indicated variable. Assume all quantities repre- sent positive real numbers.

26. Solve for h: I = bh3

12 . 27. Solve for a: I0 =

5 p

3a4

16

28. Solve for g: T = 2⇡ r

L g 29. Solve for v : L = L0

r

1 � v 2

c2 .

In Exercises 30 - 35, rationalize the numerator or denominator, and simplify.

30. 4

3 � p

2 31.

7 3 p

12x 7 32.

p x �

p c

x � c

33. p

2x + 2h + 1 � p

2x + 1 h

34. 3 p

x + 1 � 2 x � 7

35. 3 p

x + h � 3 p

x h

0.9 Radicals and Equations 125

0.9.3 Answers

1. 3|x| 2. 2t 3. 5|y 3| p

2

4. |2t + 1| 5. |w � 8| 6. p

3x + 1

7. p

c2 � v 2 |c|

8. 2r 3

p 3⇡r 2

L 9.

2"2 4 p

2⇡ |⇢3|

10. � 1 p

x 11.

3 � 6t 2 p

1 � t 2

12. 6 � 8z

3( 3 p

1 � z)2 13.

4x � 3 (2x � 1) 3

p 2x � 1

14. x = � 3 2

15. y = �1, 2 16. t = � 3 p

3 2

17. x = 5 18. t = ±3 p

7 19. x = 3

20. y = �3 21. t = � 1 3

, 2 3

22. x = 5 +

p 57

8

23. w = p

3 24. x = 6 25. x = 4

26. h = 3 r

12I b

27. a = 2 4 p

I0 4 p

5 p

3

28. g = 4⇡2L T 2

29. v = c p

L20 � L2 L0

30. 12 + 4

p 2

7 31.

7 3 p

18x 2

6x 3 32.

1 p

x + p

c

33. 2

p 2x + 2h + 1 +

p 2x + 1

34. 1

( 3 p

x + 1)2 + 2 3 p

x + 1 + 4 35.

1 ( 3 p

x + h)2 + 3 p

x + h 3 p

x + ( 3 p

x )2