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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. 6.1 Integer Powers and the Exponent Rules Section 6.1 p1

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Repeated multiplication can be expressed as a power. For example, Here, 2 is called the base and 5 is called the exponent. Notice that 2 5 is not the same as 5 2, because 2 5 = 2 × 2 × 2 × 2 × 2 = 32 but 5 2 = 5 × 5 = 25. In general, if a is any number and n is a positive integer, then we define Notice that a 1 = a, because here we have only 1 factor of a. For example, 5 1 = 5. We call a 2 the square of a and a 3 the cube of a. Section 6.1 p2

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. When we multiply powers with the same base, we can add the exponents to get a more compact form. For example, 5 2 · 5 3 = (5 · 5) · (5 · 5 · 5) = 5 2+3 = 5 5. In general, Thus, a n · a m = a n+m. Multiplying and Dividing with the Same Base Section 6.1 p3

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Example 1 Write with a single exponent: (a) q 5 · q 7 (b) 6 2 · 6 3 (c) 2 n · 2 m (d) 3 n · 3 4 (e) (x + y) 2 (x + y) 3. Solution Using the rule a n · a m = a n+m we have (a) q 5 · q 7 = q 5+7 = q 12 (b) 6 2 · 6 3 = 6 2+3 = 6 5 (c) 2 n · 2 m = 2 n+m (d) 3 n · 3 4 = 3 n+4 (e) (x + y) 2 (x + y) 3 = (x + y) 2+3 = (x + y) 5. Section 6.1 p4 Multiplying and Dividing with the Same Base (continued)

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Multiplying and Dividing Powers with the Same Base (continued) Example 2 Write as a product: (a) 5 2+a (b) x r+4 (c) y t+c (d) (z + 2) z+2. Solution Using the rule a n+m = a n · a m we have (a) 5 2+a = 5 2 · 5 a = 25 · 5 a (b) x r+4 = x r · x 4 (c) y t+c = y t · y c (d) (z + 2) z+2 = (z + 2) z · (z + 2) 2. Section 6.1 p5 Just as we applied the distributive law from left to right as well as from right to left, we can use the rule a n · a m = a n+m written from right to left as a n+m = a n · a m.

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Multiplying and Dividing Powers with the Same Base (continued) Section 6.1 p6 When we divide powers with a common base, we subtract the exponents. For example, when we divide 5 6 by 5 2, we get More generally, if n > m, Thus, if n > m.

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Using the Operations of Arithmetic to Solve Equations (continued) Since a n /a m = a n−m we have Section 6.1 p7 Example 3 Write with a single exponent: Solution

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Multiplying and Dividing Powers with the Same Base (continued) Example 4 Write as a quotient: (a) 10 2−k (b) e b−4 (c) z w−s (d) (p + q) a−b. Section 6.1 p8 Just as with the products, we can write in reverse as Solution Since a n−m = a n /a m we have

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. When we take a number written in exponential form and raise it to a power, we multiply the exponents. For example, (5 2 ) 3 = 5 2 · 5 2 · 5 2 = 5 2+2+2 = 5 2·3 = 5 6. More generally, Raising a Power to a Power Section 6.1 p9 Thus, (a m ) n = a m·n.

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Raising a Power to a Power Example 5 Write with a single exponent: (a) (q 7 ) 5 (b) (7 p ) 3 (c) (y a ) b (d) (2 x ) x (e)((x + y) 2 ) 3 (f) ((r − s) t ) z. Solution Using the rule (a m ) n = a m·n we have (a) (q 7 ) 5 = q 7·5 = q 35 (b) (7 p ) 3 = 7 3p (c) (y a ) b = y ab (d) (2 x ) x = (e) ((x + y) 2 ) 3 = (x + y) 2·3 = (x + y) 6 (f) ((r − s) t ) z = (r − s) tz. Section 6.1 p10

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Solution Using the rule a m·n = (a m ) n we have (a) 2 3·2 = (2 3 ) 2. This could also have been written as (2 2 ) 3. (b) 4 3x = (4 3 ) x, which simplifies to 64 x. This could also have been written as (4 x ) 3. (c) e 4t = (e 4 ) t. This could also have been written as (e t ) 4. (d) Section 6.1 p11 Raising a Power to a Power (continued) Example 6 Write as a power raised to a power: (a) 2 3·2 (b) 4 3x (c) e 4t (d).

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. When we multiply 5 2 · 4 2 we can change the order of the factors and rewrite it as 5 2 · 4 2 = (5 · 5) · (4 · 4) = 5 · 5 · 4 · 4 = (5 · 4) · (5 · 4) = (5 · 4) 2 = 20 2. Sometimes, we want to use this process in reverse: 10 2 = (2 · 5) 2 = 2 2 · 5 2. In general, Products and Quotients Raised to the Same Exponent Section 6.1 p12 Thus, (ab) n = a n b n.

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Products and Quotients Raised to the Same Exponent (continued) Write without parentheses: (a) (qp) 7 (b) (3x) n (c) (4ab 2 ) 3 (d) (2x 2n ) 3n. Example 7 Solution Section 6.1 p13 Using the rule (ab) n = a n b n we have (a) (qp) 7 = q 7 p 7 (b) (3x) n = 3 n x n (c) (4ab 2 ) 3 = 4 3 a 3 (b 2 ) 3 = 64a 3 b 6 (d) (2x 2n ) 3n = (2 3n )(x 2n ) 3n = (2 3 ) n (x) 2n·3n =

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Products and Quotients Raised to the Same Exponent (continued) Write with a single exponent: (a) c 4 d 4 (b) 2 n · 3 n (c) 4x 2 (d) a 4 (b + c) 4 (e) (x 2 + y 2 ) 5 (c − d) 5 Example 8 Solution Section 6.1 p14 Using the rule a n b n = (ab) n we have (a) c 4 d 4 = (cd) 4 (b) 2 n · 3 n = (2 · 3) n = 6 n (c) 4x 2 = 2 2 x 2 = (2x) 2 (d) a 4 (b + c) 4 = (a(b + c)) 4 (e) (x 2 + y 2 ) 5 (c − d) 5 = ((x 2 + y 2 )(c − d)) 5.

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Products and Quotients Raised to the Same Exponent (continued) Section 6.1 p15 Division of two powers with the same exponent works the same way as multiplication. For example, Or, reversing the process, More generally, Thus,

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Products and Quotients Raised to the Same Exponent (continued) Section 6.1 p16 Write without parentheses: Example 9cd Solution

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Products and Quotients Raised to the Same Exponent (continued) Write with a single exponent: Example 10ce Section 6.1 p17 Solution

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. We have seen that 4 5 means 4 multiplied by itself 5 times, but what is meant by 4 0, 4 −1 or 4 −2 ? We choose definitions for exponents like 0, −1, −2 that are consistent with the exponent rules. Zero and Negative Integer Exponents Section 6.1 p18 If a ≠ 0, the exponent rule for division says But so we define a 0 = 1 if a ≠ 0. (continued on next slide)

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Zero and Negative Integer Exponents (continued) Section 6.1 p19 (continued from previous slide) The same idea tells us how to define negative powers. If a ≠ 0, the exponent rule for division says But so we define a −1 = 1/ a. In general, we define Note that a negative exponent tells us to take the reciprocal of the base and change the sign of the exponent, not to make the number negative.

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Solution (a) Any nonzero number to the zero power is one, so 5 0 = 1. (b) We have (c) We have (d) We have (e) We have Section 6.1 p20 Zero and Negative Integer Exponents (continued) Evaluate: (a) 5 0 (b) 3 −2 (c) 2 −1 (d) (−2) −3 (e) Example 11

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Solution (a) We have (Solution continued on next slide.) Section 6.1 p21 Zero and Negative Integer Exponents (continued) With these definitions, we have the exponent rule for division, where n and m are integers. Rewrite with only positive exponents. Assume all variables are positive. Example 12a

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Solution (Solution continued from previous slide.) (b) We have (c) We have (d) We have Section 6.1 p22 Zero and Negative Integer Exponents (continued) Rewrite with only positive exponents. Assume all variables are positive. Example 12bcd

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Section 6.1 p23 Zero and Negative Integer Exponents (continued) In general:

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Section 6.1 p24 Zero and Negative Integer Exponents (continued) Write each of the following expressions with only positive exponents. Assume all variables are positive. Example 13 Solution

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Expressions with a Common Base If m and n are integers, 1. a n · a m = a n+m 2.3. (a m ) n = a m·n Expressions with a Common Exponent If n is an integer, 1. (ab) n = a n b n 2. Zero and Negative Exponents If a is any nonzero number and n is an integer, then: a 0 = 1 Summary of Exponent Rules Section 6.1 p25

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Be aware of the following notations that are sometimes confused: ab n = a(b n ), but, in general, ab n ≠ (ab) n, −b n = −(b n ), but, in general, − b n ≠ (−b) n, −ab n = (−a)(b n ). For example, −2 4 = −(2 4 ) = −16, but (−2) 4 = (−2)(−2)(−2)(−2) = 16. Summary of Exponent Rules (continued) Section 6.1 p26 Common Mistakes

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Evaluate the following expressions for x = −2 and y = 3: (a) (xy) 4 (b) −xy 2 (c) (x + y) 2 (d) x y (e) −4x 3 (f) −y 2. Example 14 Solution (a) (−2 · 3) 4 = (−6) 4 = (−6)(−6)(−6)(−6) = 1296. (b) −(−2) · (3) 2 = 2 · 9 = 18. (c) (−2 + 3) 2 = (1) 2 = 1. (d) (−2) 3 = (−2)(−2)(−2) = −8. (e) −4(−2) 3 = −4(−2)(−2)(−2) = 32. (f) −(3) 2 = −9. Section 6.1 p27 Summary of Exponent Rules (continued)

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Section 6.1 p28 6.1 INTEGER POWERS AND THE EXPONENT RULES Key Points Exponent rules with expressions having a common base Exponent rules with expressions having a common exponent Exponent rules with expressions having zero and negative exponents

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. 6.2 Fractional Exponents and Radical Expressions Section 6.2 p29

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. A radical expression is an expression involving roots. For example, is the positive number whose square is a. Similarly, the cube root of a, written is the number whose cube is a. In general, the n th root of a is the number b, such that b n = a. If a < 0, then the n th root of a exists when n is odd, but is not a real number when n is even. Section 6.2 p30

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Solution (a) Since 11 2 = 121, we have (b) The fifth root of −32 is the number whose fifth power is −32. Since (−2) 5 = −32, we have (c) Since the square of a real number cannot be negative, does not exist. (d) Since Section 6.2 p31 Example 1 Evaluate:

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. The laws of exponents suggest an exponential notation for roots involving fractional exponents. For instance, applying the exponent rules to the expression a 1/2, we get (a 1/2 ) 2 = a (1/2)·2 = a 1 = a. Thus, a 1/2 should be the number whose square is a, so we define Similarly, we define Using Fractional Exponents to Describe Roots Section 6.2 p32

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. The Exponent Laws Work for Fractional Exponents Example 2bd Evaluate (b) 9 −1/2 (d) 27 −1/3. Solution (b) Using the rules about negative exponents, we have (d) Since 3 3 = 27, we have Section 6.2 p33

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Example 3a Find(a) 64 2/3 (c) (−216) 2/3 (d) (−625) 3/4 Solution (a) Writing 2/3 = 2 · (1/3) we have 64 2/3 = 64 2·(1/3) = We could also do this the other way around, and write 2/3 = (1/3) · 2: 64 2/3 = 64 (1/3)·2 =(64 1/3 ) 2 = (See next slide for Example 3cd.) The Exponent Laws Work for Fractional Exponents (continued) Section 6.2 p34 Fractional Exponents with Numerators Other Than One We can also use the exponent rule (a n ) m = a nm to define the meaning of fractional exponents in which the numerator of the exponent is not 1.

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Example 3cd Find (c) (−216) 2/3 (d) (−625) 3/4 Solution (c) To find (−216) 2/3, we can first evaluate (−216) 1/3 = −6, and then square the result. This gives (−216) 2/3 = (−6) 2 = 36. (d) Writing (−625) 3/4 =((−625) 1/4 ) 3, we conclude that (−625) 3/4 is not a real number since (−625) 1/4 is an even root of a negative number. The Exponent Laws Work for Fractional Exponents (continued) Section 6.2 p35 Fractional Exponents with Numerators Other Than One (continued)

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. The Exponent Laws Work for Fractional Exponents (continued) Section 6.2 p36 Fractional Exponents with Numerators Other Than One (continued) Example 4abc Write each of the following as an equivalent expression in the form x n and give the value for n. (a) Solution (a) We haveso n = −3. (b) We have so n = 1/5. (c) We have According to the exponent laws we multiply the exponents in this expression, so (x 1/3 ) 2 = x 2/3, so n = 2/3.

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Example 4def Write each of the following as an equivalent expression in the form x n and give the value for n. The Exponent Laws Work for Fractional Exponents (continued) Section 6.2 p37 Fractional Exponents with Numerators Other Than One (continued) Solution (d) We have so n = 5/2. (e) We haveso n = −1/4. (f) We have so n = −3/2.

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Determining the Sign of an Expression from Its Factors (continued) The Exponent Laws Work for Fractional Exponents (continued) Section 6.2 p38 Example 5 Simplify each expression, assuming all variables are positive. Solution (a) Using the fact that (b) We can write

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Expressions involving roots of sums and differences will not be as easy to simplify as expressions involving roots of products and quotients. Working with Radical Expressions Section 6.2 p39 Example 6ab Do the expressions have the same value? Solution (a) We have so the expressions have different values. In general, (b) Here, In general,

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Working with Radical Expressions (continued) Section 6.2 p40 Example 6cd Do the expressions have the same value? Solution (c) We have In general, the square root of a product is equal to the product of the square roots, or (d) We have In general,

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Although the laws of exponents do not tell us how to simplify expressions involving roots of sums and differences, there are other methods that work. Again, we start with a numerical example. Simplifying Radical Expressions That Contain Sums and Differences Working with Radical Expressions (continued) Section 6.2 p41

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Section 6.2 p42 Simplifying Radical Expressions That Contain Sums and Differences (continued) Working with Radical Expressions (continued) Example 7 Are the expressions equivalent? Solution (a) We know that But, the square root of 200 cannot be 20, since 20 2 = 400, so they are not equivalent. (b) They are equivalent, since Another way to see this without evaluating is to use the distributive law to factor out the common term

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Example 8abc Combine the radicals if possible. Section 6.2 p43 Simplifying Radical Expressions That Contain Sums and Differences (continued) Working with Radical Expressions (continued) Expressing roots as fractional powers helps determine the principles for combining like terms in expressions involving roots. The principles are the same as for integer powers: we can combine terms involving the same base and the same exponent. Solution (a) We have (b) We have (c) Expressing the roots in exponential notation, we see that not all the exponents are the same:

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Example 8de Combine the radicals if possible. Section 6.2 p44 Simplifying Radical Expressions That Contain Sums and Differences (continued) Working with Radical Expressions (continued) Solution (d) Here there is a term involving and a term involving Thus the exponents are the same in both terms, but the bases are different. However, Therefore, (e) Expressing each term in exponential form, we get = 5(4x 3 ) 1/2 + 3x(36x) 1/2 = 5 · 4 1/2 (x 3 ) 1/2 + 3x(36) 1/2 x 1/2 = 10x 3/2 + 18x 3/2 = 28x 3/2.

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Rationalizing the Denominator Working with Radical Expressions (continued) Section 6.2 p45 Sometimes it works the other way around: we can simplify a fraction with a radical expression in the denominator, such as by multiplying the numerator and denominator by a carefully chosen common factor.

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Section 6.2 p46 Working with Radical Expressions (continued) Rationalizing the Denominator (continued) Example 9 Simplify by multiplying the numerator and denominator by Solution You might think that multiplying will result in a complicated expression, but in fact the result is very simple because the product is a difference of squares, and squaring undoes the square root: So multiplying the numerator and denominator of

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Section 6.2 p47 Working with Radical Expressions (continued) Rationalizing the Denominator (continued) The process in Example 9 is called rationalizing the denominator. In general, to rationalize a sum of two terms, one or more of which is a radical, we multiply it by the sum obtained by changing the sign of one of the radicals. The resulting sum is a conjugate of the original sum.

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Section 6.2 p48 Working with Radical Expressions (continued) Rationalizing the Denominator (continued) Example 10ab Multiply each expression by a conjugate. Solution (a) The conjugate of and the product is (b) The conjugate of and the product is

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Section 6.2 p49 Working with Radical Expressions (continued) Rationalizing the Denominator (continued) Example 10cd Multiply each expression by a conjugate. Solution (c) The conjugate of We rewrite the product so it is easier to see the difference of two squares: (d) Here there are two radicals, so there are two possible conjugates, We choose the first one:

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Algebra Form and Function by McCallum Connally Hughes-Hallett et al. Copyright 2010 by John Wiley & Sons. All rights reserved. Section 6.2 p50 6.2 FRACTIONAL EXPONENTS AND RADICAL EXPRESSIONS Key Points The relationship between fractional exponents and roots Exponent rules and fractional exponents Simplifying radical expressions Rationalizing the denominator

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