Roots and Radicals 145 www.petersons.com 1. ADDITION AND SUBTRACTION OF RADICALS The conditions under which radicals can be added or subtracted are much the same as the conditions for letters in an algebraic expression. The radicals act as a label, or unit, and must therefore be exactly the same. In adding or subtracting, we add or subtract the coefficients, or rational parts, and carry the radical along as a label, which does not change. Example: 2+ 3 2+ +5 3 cannot be added cannot be added2 42 22 = 92 Often, when radicals to be added or subtracted are not the same, simplification of one or more radicals will make them the same. To simplify a radical, we remove any perfect square factors from underneath the radical sign. Example: 12 4 3 2 3 27 9 3 3 3 =⋅= =⋅= If we wish to add 12 27+ , we must first simplify each one. Adding the simplified radicals gives a sum of 53 . Example: 125 20 500+ − Solution: 25 5 4 5 100 5 55 5 105 35 ⋅⋅ ⋅ = + +2 − − =− Chapter 10 146 www.petersons.com Exercise 1 Work out each problem. Circle the letter that appears before your answer. 4. Combine 1 2 180 1 3 45 2 5 20⋅+⋅−⋅ (A) 310+15+22 (B) 16 5 5 (C) 97 (D) 24 5 5 (E) none of these 5. Combine 532mn mn mn−− (A) 0 (B) 1 (C) mn (D) mn (E) − mn 1. Combine 427 248 147−+ (A) 27 3 (B) −33 (C) 93 (D) 10 3 (E) 11 3 2. Combine 80 45 20+ − (A) 95 (B) 55 (C) − 5 (D) 35 (E) −25 3. Combine 65 2 4+3 + 2−5 (A) 8 (B) 2 5+3 2 (C) 2 5+4 2 (D) 5 7 (E) 5 Roots and Radicals 147 www.petersons.com 2. MULTIPLICATION AND DIVISION OF RADICALS In multiplication and division, we again treat the radicals as we would treat letters in an algebraic expression. They are factors and must be treated as such. Example: 23 6⋅= Example: 4253 20 6⋅=⋅ Example: ()32 3232 92 18 2 =⋅=⋅= Example: 8 2 42== Example: 10 20 24 55= Example: 28 18 16 36 4 6 10()+++=== Exercise 2 Work out each problem. Circle the letter that appears before your answer. 1. Multiply and simplify: 21862⋅ (A) 72 (B) 48 (C) 12 6 (D) 86 (E) 36 2. Find 33 3 () (A) 27 3 (B) 81 3 (C) 81 (D) 93 (E) 243 3. Multiply and simplify: 1 2 26 2()+ 1 2 (A) 3 1 2 + (B) 1 2 3⋅ (C) 61+ (D) 6 1 2 + . (E) 62+ 4. Divide and simplify: 32 8 3 b b (A) 2 b (B) 2b (C) 2b (D) 2 2 b (E) bb2 5. Divide and simplify: 15 96 52 (A) 73 (B) 712 (C) 11 3 (D) 12 3 (E) 40 3 Chapter 10 148 www.petersons.com 3. SIMPLIFYING RADICALS CONTAINING A SUM OR DIFFERENCE In simplifying radicals that contain several terms under the radical sign, we must combine terms before taking the square root. Example: 16 9 25 5+ == It is not true that 16 9 16 9++= , which would be 4 + 3, or 7. Example: xx x x x x 22 2 2 2 16 25 25 16 400 9 400 3 20 − − === Exercise 3 Work out each problem. Circle the letter that appears before your answer. 1. Simplify xx 22 916 + (A) 25 144 2 x (B) 5 12 x (C) 5 12 2 x (D) x 7 (E) 7 12 x 2. Simplify 36 64 22 yx+ (A) 6y + 8x (B) 10xy (C) 6y 2 + 8x 2 (D) 10x 2 y 2 (E) cannot be done 3. Simplify xx 22 64 100 − (A) x 40 (B) − x 2 (C) x 2 (D) 3 40 x (E) 3 80 x 4. Simplify yy 22 218 − (A) 2 3 y (B) y 5 (C) 10 3 y (D) y 3 6 (E) cannot be done 5. ab 22 + is equal to (A) a + b (B) a – b (C) ab 22 + (D) (a + b) (a - b) (E) none of these Roots and Radicals 149 www.petersons.com 4. FINDING THE SQUARE ROOT OF A NUMBER In finding the square root of a number, the first step is to pair off the digits in the square root sign in each direction from the decimal point. If there is an odd number of digits before the decimal point, insert a zero at the beginning of the number in order to pair digits. If there is an odd number of digits after the decimal point, add a zero at the end. It should be clearly understood that these zeros are place holders only and in no way change the value of the number. Every pair of numbers in the radical sign gives one digit of the square root. Example: Find the number of digits in the square root of 328,329. Solution: Pair the numbers beginning at the decimal point. 32 83 29 . Each pair will give one digit in the square root. Therefore the square root of 328,329 has three digits. If we were asked to find the square root of 328,329, we would look among the multiple-choice answers for a three-digit number. If there were more than one, we would have to use additional criteria for selection. Since our number ends in 9, its square root must end in a digit that, when multiplied by itself, ends in 9. Going through the digits from 0 to 9, this could be 3 (3 · 3 = 9) or 7 (7 · 7 = 49). Only one of these would appear among the choices, as this examination will not call for extensive computation, but rather for sound mathematical reasoning. Example: The square root of 4624 is exactly (A) 64 (B) 65 (C) 66 (D) 67 (E) 68 Solution: Since all choices contain two digits, we must reason using the last digit. It must be a number that, when multiplied by itself, will end in 4. Among the choices, the only possibility is 68 as 64 2 will end in 6, 65 2 will end in 5, 66 2 in 6, and 67 2 in 9. Chapter 10 150 www.petersons.com Exercise 4 Work out each problem. Circle the letter that appears before your answer. 4. The square root of 25.6036 is exactly (A) 5.6 (B) 5.06 (C) 5.006 (D) 5.0006 (E) 5.00006 5. Which of the following square roots can be found exactly? (A) .4 (B) .9 (C) .09 (D) .02 (E) .025 1. The square root of 17,689 is exactly (A) 131 (B) 132 (C) 133 (D) 134 (E) 136 2. The number of digits in the square root of 64,048,009 is (A) 4 (B) 5 (C) 6 (D) 7 (E) 8 3. The square root of 222.01 is exactly (A) 14.3 (B) 14.4 (C) 14.6 (D) 14.8 (E) 14.9 Roots and Radicals 151 www.petersons.com RETEST Work out each problem. Circle the letter that appears before your answer. 6. Find a b a b 2 2 2 2 + (A) a b 2 2 (B) a b (C) 2a b (D) a b 2 (E) a b 2 2 7. The square root of 213.16 is exactly (A) 14.2 (B) 14.3 (C) 14.8 (D) 14.9 (E) 14.6 8. The number of digits in the square root of 14,161 is (A) 5 (B) 4 (C) 3 (D) 2 (E) 6 9. ()23 5 is equal to (A) 32 3 (B) 288 3 (C) 10 3 (D) 90 3 (E) 16 3 10. Find 25 36 4 64 16 m cd (A) 5 6 2 84 m cd (B) 5 6 2 32 4 m cd (C) 5 6 2 32 8 m cd (D) 5 6 2 88 m cd (E) 5 6 16 4 m cd 1. The sum of 28450, , and 318 is (A) 33 6 (B) 976 (C) 33 2 (D) 135 6 (E) 136 2 2. The difference between 1 2 180 and 2 5 20 is (A) 1 10 160 (B) 16 2 5 5 (C) 16 2 5 (D) 11 5 5 (E) 2 5 5 3. The product of ax2 and xx6 is (A) 23 2 ax (B) 12ax 3 (C) ()23 2 ax (D) 12ax 2 (E) 12ax 4. Divide 42 40 36 rt by 35 2 rt (A) 56 2 2 rt (B) 28 2rt rt (C) 28 2 2 rt (D) 28 2rt t (E) 56 2rt t 5. Solve for x: 3 09 x = . (A) 10 (B) 1 (C) .1 (D) .01 (E) 1.1 Chapter 10 152 www.petersons.com SOLUTIONS TO PRACTICE EXERCISES Exercise 1 1. (E) 427 49 3 12 3=⋅= 248 216 383 147 49 3 7 3 12 3 8 3 7 3 11 3 =⋅= =⋅= =−+ 2. (B) 80 16 5 4 5=⋅= 45 9 5 3 5 20 4 5 2 5 45 55 =⋅= =⋅= =+3 5−2 5 3. (C) Only terms with the same radical may be combined. 65 45 25 32 42 − +2 = = Therefore we have 25 42+ 4. (B) 1 2 180 1 2 36 5 3 5⋅=⋅⋅= 1 3 45 1 3 95 5 2 5 20 2 5 45 4 5 5 35 5 4 5 545 ⋅=⋅⋅= ⋅=⋅⋅= +−⋅= − 44 5 5 3 1 5 5 16 5 5== 5. (A) 550mn mn−= Diagnostic Test 1. (B) 75 25 3 5 3=⋅= 12 4 3 2 3 53 73 =⋅= =+2 3 2. (B) 125 25 5 5 5=⋅= 45 9 5 3 5 55 35 25 =⋅= =− 3. (D) 94 36 6 2 xx x x⋅= = 4. (B) 16 4= 2 4 24 5 x x x x = = = . . Multiply by . 5. (C) Since the last digit is 6, the square root must end in 4 or 6. 6. (B) Since the number has two digits to the right of the decimal point, its square root will have one digit to the right of the decimal point. 7. (E) 25 36 900 61 900 61 30 22 2 xx x x + == 8. (E) It is not possible to find the square root of separate terms. 9. (C) 812 23 44 42 8==⋅= 10. (D) ()()222= . Therefore, ()()2222242⋅ ⋅⋅⋅= Roots and Radicals 153 www.petersons.com Exercise 2 1. (A) 21862 123612672⋅= =⋅= 2. (B) 333333 2733 81 3⋅⋅= =⋅() 3. (A) Using the distributive law, we have 1 2 12 2 1 2 43 3+ 1 4 + 1 2 + 1 2 ⋅= ⋅ = 4. (C) Dividing the numbers in the radical sign, we have 42 2 bb= 5. (D) 348 316 3 123=⋅= Exercise 3 1. (B) 16 9 25 144 5 12 22 2 xx x x+ 144 == 2. (E) The terms cannot be combined and it is not possible to take the square root of separated terms. 3. (D) 100 64 6400 36 6400 6 80 3 40 22 2 xx xxx- === 4. (A) 18 2 16 36 4 6 2 3 22 2 yy y yy- 36 === 5. (E) It is not possible to find the square root of separate terms. Chapter 10 154 www.petersons.com Exercise 4 1. (C) Since the last digit is 9, the square root must end in 3 or 7. 2. (A) Every pair of digits in the given number gives one digit of the square root. 3. (E) Since the number ends in 1, its square root must end in 1 or 9. 4. (B) Since the number has four digits to the right of the decimal point, its square root will have two digits to the right of the decimal point. 5. (C) In order to take the square root of a decimal, it must have an even number of decimal places so that its square root will have exactly half as many. In addition to this, the digits must form a perfect square (. .)09 3= . Retest 1. (C) 28 24 2 42=⋅= 450 425 2 202 318 39 2 9 2 42 202 92 332 =⋅= =⋅= =++ 2. (D) 1 2 180 1 2 36 5 3 5=⋅= 2 5 20 2 5 45 4 5 5 35 4 5 5 11 5 5 =⋅= −= 3. (A) axxxax xax26 1223 22 ⋅= = 4. (C) 42 40 35 14 8 14 8 28 2 36 2 24 24 2 rt rt rt rt rt = = 5. (A) 09 3= 3 3 33 10 x x x x = = = . Multiply by 6. (D) 22 2 2 a b a b = 7. (E) Since the last digit is 6, the square root must end in 4 or 6. 8. (C) A five-digit number has a three-digit square root. 9. (B) 2323232323 3293 2883⋅⋅⋅⋅= =() 10. (C) 25 36 5 6 4 64 16 2 32 8 m cd m cd = [...]... or parts of terms We may only simplify by dividing factors Exercise 4 Work out each problem Circle the letter that appears before your answer 1 1 3 − 5 2 Simplify 3 4 15 (A) 26 15 (B) − 26 (C) (D) (E) 2 2 26 15 26 − 15 a x2 Simplify a 2 x x (A) a 1 (B) 3 1 1 − x y Simplify 1 1 + x y x−y (A) x + y x+y (B) x − y y− x (C) x+y (D) (E) 4 1 –xy 1 Simplify 1 + x 1 y x+y (A) x (B) (C) 2y x +1 y +1 x x +1 y... 32 (C) 4 (D) 36 (E) 6 8 If (x + y)2 = 10 0 and xy = 20, find x2 + y2 (A) (B) (C) (D) (E) 10 0 20 40 60 80 www.petersons.com 9 1 1 1 1 1 1 1 1 If x + y = 2 and x − y = 4 , find x 2 − y 2 (A) (B) (C) (D) (E) 3 4 1 4 3 16 1 8 7 8 10 The trinomial x2 – x – 20 is exactly divisible by (A) x – 4 (B) x – 10 (C) x + 4 (D) x – 2 (E) x + 5 Factoring and Algebraic Fractions 1 SIMPLIFYING FRACTIONS In simplifying... (D) (E) a b 5 27 x 5 3y 9x6 y3 9x5 y3 27 x 5 y3 27 x 6 y3 1 Simplify 2 + a b a 2a + 1 (A) b (B) 0 15 5 4a2 + 1 xy (E) ( x − 5 )2 ( x + 5 )2 ( x − 5 )2 − ( x + 5 )2 2a + 1 ab (D) 1 1 2a + 1 a (C) x−5 5− x by x+5 5+ x 3 Find an expression equivalent to  y    2b + 1 b 15 6 Chapter 11 6 1 1 − a b Simplify 2 b−a (A) 2 a−b (B) 2 b−a (C) 2ab ba (D) 2 2ab (E) b+a 7 If x + y = 16 and x2 – y2 = 48, then... = (A) (B) (C) (D) (E) 1 12 1 7 2 7 1 6 1 1 and a − b = , find a2 – b2 3 4 none of these 2 If (a – b)2 = 40 and ab = 8, find a2 + b2 (A) 5 (B) 24 (C) 48 (D) 56 (E) 32 3 If a + b = 8 and a2 – b2 = 24, then a – b = (A) 16 (B) 4 (C) 3 (D) 32 (E) 6 www.petersons.com 4 The trinomial x2 + 4x – 45 is exactly divisible by (A) x + 9 (B) x – 9 (C) x + 5 (D) x + 15 (E) x – 3 5 If 1 1 1 1 1 1 − = 5 and + = 3 , then... factoring of 1 www.petersons.com 15 7 15 8 Chapter 11 Exercise 1 Work out each problem Circle the letter that appears before your answer 1 3x 3 − 3x 2 y Simplify to simplest form: 9 x 2 − 9 xy x (A) 6 x (B) 3 2x (C) 3 (D) (E) 2 (B) (C) (D) (E) 3 2x − 8 12 − 3x 0 1 1 www.petersons.com 4 3 b+4 b+5 b−4 b−5 b+4 − b+5 2x + 4 y Simplify to simplest form: 6 x + 12 y (A) 2 3 (B) − (C) (D) 3x − y 2 b 2 + b − 12 b 2... resulting fractions, giving an answer of xy 1 1 Example: Divide 15 a 2 b by 5a3 2 Solution: We invert the divisor and multiply 15 a 2 b 1 ⋅ 3 2 5a We can divide the first numerator and second denominator by 5a2, giving 3b 1 3b ⋅ or 2 a 2a www.petersons.com 16 1 16 2 Chapter 11 Exercise 3 Work out each problem Circle the letter that appears before your answer 1 x2 y4 Find the product of 3 and 5 y x y2 (A)... number in the numerator and denominator, the value of 9 3 9 3⋅3 = to , we are really saying that and then 12 4 12 3 ⋅ 4 9 5+ 4 9 4 = = This is dividing the numerator and denominator by 3 We may not say that and then say that 12 5 + 7 12 7 9t 3 9+ t = because we divide numerator and denominator by 3t However, a serious error in algebra as well 12 t 4 12 + t the fraction would not remain the same When we... www.petersons.com 15 9 16 0 Chapter 11 Exercise 2 Work out each problem Circle the letter that appears before your answer 1 Subtract (A) (B) (C) (D) (E) 2 Add (A) (B) (C) 6x + 5y 4 x + y − 2x 2x Add 5 Subtract 1 + 4y 4y 1 + 2y x + 2y x x + 3y x 3c 3d + c+d c+d 6cd c+d 3cd c+d 3 2 x+4 1 + 6 2 x+7 (A) 6 x +5 (B) 8 x+4 (C) 12 x +5 (D) 12 x +5 (E) 6 4 (A) (B) − 2b 3 b 5 b 20 (D) (C) (E) 3 3 9cd c+d (D) b (E)... Simplify 1 + x 1 y x+y (A) x (B) (C) 2y x +1 y +1 x x +1 y a2 x 1 ax (D) (D) ax (E) a x 1 Simplify 2 + t 2 t2 (C) (E) 5 (A) (B) t2 + t t3 (C) 2t + 1 2 (D) t +1 (E) 4+t 2 www.petersons.com 16 3 16 4 Chapter 11 5 USING FACTORING TO FIND MISSING VALUES Certain types of problems may involve the ability to factor in order to evaluate a given expression In particular, you should be able to factor the difference of.. .11 Factoring and Algebraic Fractions DIAGNOSTIC TEST Directions: Work out each problem Circle the letter that appears before your answer Answers are at the end of the chapter 1 Find the sum of (A) (B) (C) (D) (E) 2 (B) (C) (D) (E) 2−a b 2−a 2−b a − 2b b 2b − a b 2a − b b Divide (A) (B) (C) (D) (E) 4 2n 2 7 3n 7 11 n 12 2n 2 12 9n 12 Combine into a single fraction: . 1 (C) .1 (D) . 01 (E) 1. 1 Chapter 10 15 2 www.petersons.com SOLUTIONS TO PRACTICE EXERCISES Exercise 1 1. (E) 427 49 3 12 3=⋅= 248 216 383 14 7 49 3 7 3 12 3 8 3 7 3 11 3 =⋅= =⋅= =−+ 2. (B) 80 16 5. a b a b 2 2 2 2 + (A) a b 2 2 (B) a b (C) 2a b (D) a b 2 (E) a b 2 2 7. The square root of 213 .16 is exactly (A) 14 .2 (B) 14 .3 (C) 14 .8 (D) 14 .9 (E) 14 .6 8. The number of digits in the square root of 14 ,16 1 is (A) 5 (B) 4 (C) 3 (D) 2 (E) 6 9. ()23 5 is equal to (A). 28450, , and 318 is (A) 33 6 (B) 97 6 (C) 33 2 (D) 13 5 6 (E) 13 6 2 2. The difference between 1 2 18 0 and 2 5 20 is (A) 1 10 16 0 (B) 16 2 5 5 (C) 16 2 5 (D) 11 5 5 (E) 2 5 5 3. The product of