Radicals. Stephen Perencevich

Transcription

1 Radicals Stephen Perencevich

2 Stephen Perencevich Georg Cantor Institute for Mathematical Studies Silver Spring, MD c 009 All rights reserved.

3 Algebra II: Radicals 0 Introduction Perencevich Section 0: Introduction We have already extensively used the square root, mainly in connection with the solution of quadratic equations. The square root is the simplest example of a radical. In this section we will study other radicals and see that they obey the same rules as exponents, and, in fact, that radicals are a kind of exponent. We begin by defining radical notation and giving some examples. Definition For a positive integer n, n a = x if and only if the following two conditions are satisfied. (i) x n = a (ii) a and x have the same sign. Condition (ii) is somewhat of a technical matter. We will illustrate both conditions with examples. We employ the usual notational convention that a = a. Example Find 4 6. Since 4 = 6, 4 6 =. Note that ( ) 4 = 6, but by condition (ii) 6= 4 6. We now make the following observation. By definition 4 6 = x if and only if x 4 = 6 and x is positive. Another way to describe 4 6 is to say that it is the positive root of the polynomial equation x 4 6 = 0. The diagram to the left shows the graph of y = x 4 6. There are two x-intercepts, and. These correspond to the real roots of this polynomial. Since this is a degree-four polynomial, it must, therefore, have two complex roots. Exercise Show that (i) 4 = 6 and ( i) 4 = 6. 4 Example Find 8. Since ( ) = 8, 8 =. We now make the following observation. By definition 8 = x if and only if x = 8 and x is negative. Another way to describe 8 is to say that it is the negative root of the polynomial equation x + 8 = 0. The diagram to the left shows the graph of y = x + 8. There is one x-intercept,. This corresponds to the only real root of this polynomial. Since this is a degree-three polynomial, it must, therefore, have two complex roots. 5 Exercise Show that ( + i ) = 8. Can you say what the other complex root is?

4 Algebra II: Radicals 0 Introduction Perencevich We mentioned in the last example that ( ) = 8. Let us calculate this result explicitly for clarity. ( ) = ( )( )( ) = (+4)( ) = 8 It is clear that a negative number raised to an odd power will be negative. The last two examples illustrate the fact that n a is the positive root of the polynomial y = x n a if a > 0 and the negative and only real root of y = x n a if a < 0. This is an interesting fact and one might wonder what can be said about the full set of roots of these special polynomials. This topic is taken up more fully in precalculus courses. For our purposes finding n th roots of numbers amounts to raising numbers to powers as the next example shows. 6 Example Find the following values. a) 4 b) 5 04 a) We need to find x so that x = 4. Since 4 is odd, we need only check odd values of x. We proceed by guess and check. = 7, 5 = 5, 7 = 4 so 4 = 7 b) We need to find x so that x 5 = 04. Since 04 is even, we need only check even values of x. We proceed by guess and check. so 5 04 = 4 Problems A Find the following values. 5 =, 4 5 = B Simplify the following expressions. Use only positive exponents.. µ a b c a b 5 c 6 4. x y x y 7. µ a b c a d 5 4. x y x y. µ a b c 5. a b 4 c 5 6. (m n p ) (m n p 4 ) µ z µ 9y 8. ( 4) 9. (a x y ) y z C Factor the following expressions completely.. x 9. a 5 6a. m r b 5. x x x + 4x + 7. a b d 6 4ab 7 c 8. x 4 x m 6h

5 Algebra II: Radicals Rational Exponents Perencevich Section : Rational Exponents Radicals can be expressed as exponents as the following definition indicates. Definition For positive integers m and n (i) a /n = n x (ii) a m/n = n a m = n a m We add the two parts of this definition to our list of exponent rules. Rules for Exponents. a m a n = a m+n 4. a m a n = am n 6. a n = a n (a 6= 0) 8. n a = a /n. (a m ) n = a m n. (a b) m = a m b m 5. a b m = a m b m 7. a 0 = (a 6= 0) 9. n am = a m/n Rules 6 apply for rational as well as integer exponents. Example a) 44 / = 44 = b) 5 / = 5 = 5 c) 8 / = 8 = = 4 d) 6 /4 = 4 6 = = 8 Since rule applies to rational exponents we have the following: n a b = n a n b. This allows us to simplify other radicals as we do square roots. For example = 4 =. Example Simplify the following expressions: a) x 6 b) 6 c) x 9 d) x 4 a) x 6 = x 6/ = x To take the cube root of x 6 we simply divide the exponent 6 by. b) 6 = 8 = c) x 9 = x 8 x = x 8/ x = x 4 x d) x 4 = x x = x / x = x 4 x 4 Exercise Simplify the following expressions: a) 4 5 b) 4 6x c) x 7 d) x 7 6 is not a perfect cube, so we look for a perfect cube that divides evenly into 6 and 6 = 8. x9 = x 9/. Since 9 is not divisible by, we factor x 9 as follows: x 9 = x 8 x. x 4 = x 4/. Since 4 is not divisible by, we factor x 4 as follows: x x Ans: a) 6 b) x c) x 8 x d) x 5 x

6 Algebra II: Radicals Rational Exponents Perencevich We can use the rules of exponents to simplify complex expressions involving radicals and rational exponents. 5 Example Simplify π π π / π π π / = π π / π / = π7/ π / = π5/ = π 5 = π π = π π 6 Example Simplify r m m r m m = m / m / µ m m / m / / = m m/6 m / = m7/6 m / = m7/6 / = m / = m 7 Exercise Simplify the following expressions. Express your answers with radicals, not rational exponents. a) e4 4 e e /4 b) 4 s k k k 4/ Ans: a) e e b) 4 k 8 Example Simplify r 8 x First we use rule 5. 8 =, so we can completely simplify the numerator. We use rule 9 on the denominator. r 8 x = 8 x r 8 x = x = / x 7 9 Example Simplify 5 x 8 / First we use rules and. 5 x 8 / = 5 / x 8 / = 5 / x 4 Now we use rules 6 and 8. 5 / x 4 = x4 x4 = = x4 5/ 5 5 µ a 6 b 9 0 Example Simplify µ a 6 b 9 c 8 c 8 / / = a4 b 6 c = a4 c b 6 Example Simplify q 64a q 64a = 64a / / = 64a /6 = 64 /6 a = 6 64 a = a 4

7 Algebra II: Radicals Rational Exponents Perencevich The next examples illustrate how to use the rules of exponents to solve equations. Example Solve for n: p n Q 4 = Q 6. Use exponent rules to convert the radical to a rational exponent. Set the exponents equal. Q 4/n = Q 6 4 n = 6 n = 4 Example Solve for n: b /n = b 8. Use exponent rules to simplify the left side. Set the exponents equal. b /n = b 8 4 n = 8 n = b 4/n = b 8 4 Example Solve for n: a / a n/6 = a 5/. Use exponent rules to simplify the left side. 5 Example Solve for n: Set the exponents equal. q k k = n k 6. Use exponent rules to simplify the left side. Rewrite the right side. k 6/n a (4+n)/6 = a 5/ The equation now becomes: k / = k 6/n Set the exponents equal. To clear fractions, multiply both sides by n 4 + n = n = 5 n =. q k k = k k / / = k 4/ / = k / = 6 n (n) = 6 n n = 8 n = 9. 5

8 Algebra II: Radicals Rational Exponents Perencevich Problems A Simplify the following expressions / / 6. / / / x. 7a 6. a b. 5x 4. b 5 z 5. a 6 b 8 / r r r 7 4 y x z 0 9. a b 9 c x q. p p q p 9 4. q p p 5 q 6. k 5 p k 0 4. m 4 4 m 8 π π 5. π / π 4 s z 5 µ z 7a b µ 8m 9 n 6 6. x x x 7/ 7. r r r 7 8. z 5/ 9. c / 0. B Simplify the following expressions. Express your answers with radicals only, not rational exponents (x 4 ) /. 40. (a b) / (a 6 b 9 c ) / 5. x 7 6. (6a 5 x 8 ) / 7. x 8. x 5 9. a 0. a b 7. 5x 7. b 5 z /. a 6 b 8 / p 4. 8p6 q 7 r 8 5. p x 4 y a 6 b 8 c 9 7. a b 7 c 8 8. a b 9 x 4/ 9. 8x 0. 5x 9 /. q 4x 5. p p p π q π. π 5/ 4. m 5 m 5. (a b 4 c ) /4 (ab 6 c 4 ) / π 6 s z 5/ z 7 6. k k 7 k 5 7. C Find the value of n. x 4 x 4 x 9 8. z k 8 / 9. b b 6 b 40. p 4 q 7 / p q 5 / n. a n = a 4. b 44 n = b. x 7 = x 5 x / n 4. (a 6 ) /n = a 4 5. (a 0 ) n/ = a 5 6. b = b n 7. a /4 a n/6 = a 9/ 8. a6 = 4 a 4 9. a = a n n 0. a 5 = a 5. an = a 4. n 64 = 4. n 4 = 4. x n x n = x 4 5. w = w 6 6. p m n p m = m / 4 7. k n q k = k 4/ n 8. p p p = p / D Factor completely the following expressions.. x y 64. z 64. x 5x + 6x 4. x 4 5. ax 49a a b 7. x 5x 7 8. ax a b a + 5b 0. x 4 6x + 9x. x 4 + x 5 m 7n 6

9 Algebra II: Radicals Radical Arithmetic Perencevich Section : Radical Arithmetic Radicals in expressions can be treated much like variables. Example Simplify the the following expressions by combining like terms. a) (x + 5y + ) + (x 7y + 8) b) These problems are quite similar. We will solve them side-by-side. a) (x + 5y + ) + (x 7y + 8) b) First we rearrange the terms to put like terms near each other. x + x + 5y 7y Now combine like terms. 4x y In some cases, radicals that seem not to be alike turn out to be alike after simplification. Example Simplify + 8. The radicals and 8 seem not to be alike, but since 8 =, they are alike after simplification. + 8 = + = + 4 = 7. Similar problems involving variables can also be handled in this way. Example Simplify x 7 + 5x x 5. Before we can begin, we will simplify the radicals involved. x7 = x 7 = x + = x x = x x x5 = x 5 = x + = x x = x x x 7 + 5x x 5 = x x + 5x x x = x x + 5x x = 7x x. 4 Exercise Simplify the following expressions. a) 4 b) x 9 + 4x x 7 5 Example Simplify 8 x + x 8 First we simplify both terms. 8 x = 8 x = x = x x 8 = x = x 8 x + x 8 = x + x = 5x 6 Example Simplify x 5 + 4x 7x x 5 = x 5 = 4 x 4 x 4x 7x = 4x 7 x = 4x 9 x x = x 5 + 4x 7x = 6x x + x x = 8x x 7 x x = 6x x = 6x x = 4x x x = x x = x x

10 Algebra II: Radicals Radical Arithmetic Perencevich We can also multiply expressions involving radicals. 7 Example Multiply the following expressions. a) b) x x + c) x 5 x + 5 These problems are handled by binomial multiplication. a) = (5)() = b) x x + = x + x x = x c) x 5 x + 5 = x x + 5 x = x x x 5 x + x = x x 4 Recall that in most cases it is preferable to have a rational number for the denominator of a fraction. These same techniques of radical arithmetic are used to rationalize the denominator of a fraction. 8 Example Rationalize the denominators of the following fractions. a) b) c) a) b) c) = = + 7 = + 9 = = = = = 6 8

11 Algebra II: Radicals Radical Arithmetic Perencevich Problems A Simplify the following expressions x + 8x 9. x 5 7x 5 0. x x + x 5. x 7 + x x 5. 8x 9 + x 9. x 7 x x 5 4. x + x x x 8x 6. x + x x 7. x 7 + x 7x 5 8. x 0x 7 45x 9 B Simplify the following expressions x + 7 x 7. x 0 x + 0. x + 8 x 8. x + x 4. x + x 5. x 8 x x + x 7. x x + 8. x + 5 x 5 C Simplify the following expressions.. 54 /.. (7x 7 ) / (6x ) / (5a 6 b 8 ) / 8. x 9. x 0. x 5. a. a b 7. 5x 7 / 4. b 5 z 5. r a 6 b 8 / 7 6. x µ y 8 /4 r z 0 9. a b 7 c 8 / x /. 9x 7. a 6 b c / a b 5 c 4 / 4. 8x D Rationalize the denominators of the following fractions E Solve the following equations.. x(5 x) = 6. x(x + 7) = 6( + x). (x 4)(x + 4) = (x + ) 4. (x + 5)(x + 5) = 4 5. (x )(x + ) = x + 6. x(x 7) = 8 7. x(x ) = ( x) 8. (x 6)(x + 6) = (x + ) 9. (x + )(x + 5) = x 0 9

12 Algebra II: Radicals Radical Equations Perencevich Section : Radical Equations In this section, we will learn how to solve equations involving radicals. Like with rational equations, the method of solution of radical equations sometimes introduces extraneous roots. The reason why is that the equation a = b does not imply a = b. This is easy to see since ( ) = (), but 6=. The fact that extraneous roots sometimes occur does not present a substantial problem, but it does necessitate the need to check solutions. Let us proceed immediately to the examples. Example Solve x + = x. We proceed by squaring both sides of the equation. x + = x x + = x The result is a quadratic equation that can be solved by factoring. x x = 0 (x )(x + ) = 0 x = x = The apparent solution set is {, }, but we need to check these solutions. x + = x x = x = + = + = = False = True is extraneous, but is a true root, so the solution set is {}. Example Solve 7x + 7 = x. We wish to square the equation to obtain an equation without a radical. Before squaring, however, we need to isolate the radical by adding to both sides. Now we square both sides of the equation. 7x + 7 = x + 7x + 7 = (x + ) = (x + )(x + ) 7x + 7 = x + x + The result is a quadratic equation that can be solved by factoring. x 5x 6 = 0 (x + )(x 6) = 0 x = x = 6 The apparent solution set is {, 6}, but we need to check these solutions. 7x + 7 = x x = x = 6 p p 7( ) + 7 = 7(6) + 7 = 6 0 = True 7 = 6 True In this case both are true roots, so the solution set is {, 6}. 0

13 Algebra II: Radicals Radical Equations Perencevich Example Solve 4 x + = x + 6. We begin by squaring both sides of the equation. The result is a quadratic equation that can be solved by factoring. 4 x + = (x + 6) = (x + 6)(x + 6) 6(x + ) = x + x + 6 6x + = x + x + 6 x 4x + 4 = 0 (x )(x ) = 0 x = x = The apparent solution set is {}, but we need to check this solution. x = 4 + = + 6 4() = 8 False In this case the only possible root,, is extraneous so the solution set is { }, the empty set. 4 Example Solve x + = 6x. We begin by squaring both sides of the equation. The result is a quadratic equation that can be solved by factoring. x + = ( 6x) x + = 6x 6x x + = 0 (9x + )(4x ) = 0 x = 9 x = 4 The apparent solution set is Ω 9, æ, but we need to check these solutions. 4 x + = 6x x = /9 x = /4 r 9 µ + = 6 r µ = 6 4 r µ 6 9 = 6 r µ = = 4 True = False In this case Ω 4 is extraneous and 9 is a true root, so the solution set is æ. 9 5 Example Solve x + + x = 4. When an equation has two radicals the solution will require two squaring processes, but first one of the radical expressions must be isolated.

14 Algebra II: Radicals Radical Equations Perencevich We begin by isolating one of the radicals. x + + x = 4 x + = 4 x We then square both sides of the equation. Rearranging. We now square again. The result is a quadratic equation that can be solved by factoring. x + = (4 x) x + = 6 6 x + 4x x 5 = 6 x ( x 5) = ( 6 x) x 6x + 5 = 0 (x )(x 5) = 0 x = x = 5 The apparent solution set is {, 5}, but we need to check these solutions. x + + x = 4 x = x = 5 p p () + + = 4 (5) = = 4 4 = 4 True (5) = = 4 False In this case 5 is extraneous and is a true root, so the solution set is {}.

15 Algebra II: Radicals Radical Equations Perencevich Problems A Solve the following equations.. + x + x =. + x + x =. x + + x = 4 4. x + = x x + 6 x = 6. 4 x + = 4x x x = 4 8. x + 6 = x 9. x + x + 6 = 0. 9x + 9 = x x x = x x = 6. + x x = x x = 4 5. x + + x = 4 6. x + + x = 7. x + = x + 8. x + = (x + ) 9. x + + x = 5 0. x + + x =. x 4 x = 6. x + + x =. x + + x = 4. x + x = B Simplify the following expressions. Express your answers using radicals, not rational exponents x x 6. 7 a b 6 c a 6 b 9 c / 6a 8 c /4 9. a 7 r 5 m 7 C Simplify the following expressions x 7 + x x 5 5. p p p + p p 5 6. m m + 5 m 7 7. x7 + x x 4 8. p p 8p + 4 p 8p 5 9. m m + 7m 7 D Find the value of n.. x n/ x /6 = x 5/6. 4. n a = a 6 7. a / a n/4 = qa 4 4 a 8. E Solve the following equations. p n p 4 = p m 6 /n = m 9 p n q 0 = q 5/ 9. n a = a 8 6. y y n = y 5 q m = n m. x + = 7. x = 7. 4 x = 4. 5x + = 0 5. x + 5 = 6. x + =

16 Algebra II: Radicals Chapter Review Perencevich Chapter Review A Simplify the following expressions. Express your answers with radicals only, not rational exponents p 7. 6 a4 b 8 c b5 x 0 y 5 9. x 0. 7m n 6 p /. 6q 4 r z /4. m 5 r 0 p 5 /5 r p. 7y 5 4. m 5. a 6 b 8 / µ / x 8 7. a9 b 7 c 8 8. (a b 9 c ) /9 (a b 6 c ) / B Find the value of n. n. a n = a. b 44 = b (a 9 ) /n = a 6 5. (a 0 ) n/ = a 0 6. n 7. a /4 a n/ = a / 8. a = a 4 9. n 0. a 5 = a 5. 6 an = a 4.. n 4 = x x n = x 5 5. n x 6 = x 5 x n b 5 = b 4 a = 6 a n n 5 = 5 n w 8 = w C Simplify the following expressions x 8x + 8x x 5 7x 5 0. x x + 8x 5. 7x 7 + x x 5. 8x x 9 D Simplify the following expressions x + x 8. x x + 9. x + 6 x 6 E Rationalize the denominators of the following fractions F Solve each of the following equations.. 4x + 7 = x +. x + 4 = x. 0x + 0 x = 4. x 4 = x x = 5 + x + 6. x = + x x + = x x + = x x + x = 4

17 Algebra II: Radicals Chapter Review Perencevich Chapter Review Answers A ab c 0 8. bx y 5 9. x 5 x 0. 9m n 4 p 8. 8q r 9 z 9. 4m r 4 p 6. y p y 5 4. m 5. a 4 b 6. x 7 7. a b c bc 8. ab c C x x 9. x x 0. 7x x. 5x x. 7x 4 x B D x x 8. x 4x + 9. x 6x + E F. Ω, æ. {}. Ω, 9 æ {0} 5. { } 6. {6} 7. {0} 8. {, } 9. {4} 5

18 Algebra II: Radicals Answers to Odd-Numbered Problems Perencevich Section 0 A B. 7. c 0 a 8 b 4. 8 c 4 d 0 6 a 0 b 9. m 9 p 6 n 4 5. a 8 b c x 4 a 6 y 4 C. (x )(x + ). (m 4)(m + 4m + 6) 5. (x + )(x 7) 7. ab (ad 6 b 5 c ) 9. (5m 4h)(5m + 4h) Section A a. 5x 7 5. a 4 b 7. y 9. ab c 5. p. k 5. π C b 4 ac 7. n =. n = 5. n = 7. n = 5 9. n =. n =. n = 5. n = 4 7. n = 5 B x x 7. x 0 x. 5x x. a 4 b 5. x y y 7. ab c bc 9. x 5 x. x 6x. 5. ab 4 c c 7. x x 9. b D. (xy 8)(xy + 8). x(x )(x ) 5. a(x 7)(x + 7) 7. (x 7)(x + ) 9. (a + 5b)(9a 5ab + 5b ). (x + 5)(x ) π Section A x x. x x. x x 5. x 7. 8x x B x 40. 4x 5. x 8 7. x 4x + C. 6. x x 5. x 7. 5a 4 b 9. x 0 x. a 7 a. 5x x 5. a 4 b 7. y. 7. D ab 4 c 5 b c. 4x 5 x. a 5 b c E. {, }. { 5, } 5. {, 5} 7. {, } 9. { 0, } 6

19 Algebra II: Radicals Answers to Odd-Numbered Problems Perencevich Section A. {}. { } 5. {, 5} Ω 7. {} 9., 5 æ. {4, 5} Ω æ 5. 9 Ω æ {7} 7. {}. Ω æ 5 9 B x 7. ab c m 9 C p p 7. x x D. n =. n = 5. n = 7. n = 9. n = 9 E. {8}. Ω 6, 6 æ 5. { } 9. m m 7