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Exponents and Radicals are the topic for this months provincial exam practice questions. These topics are probably familiar to most of you but a little refresher never hurts. We will start out by reviewing a few helpful rules. Radicals: are any number that can be expressed in the form radicals you will experience are listed below. √ n x. The most common types of √ Square Roots - x means that we are seeking √ to find a whole number with the property that multiplied by itself it yields x. Example: 81 = 9 since 9 × 9 = 81. √ Cube Roots - 3 x means that we are seeking to find a whole number with the property that √ 3 multiplied by itself twice it yields x. Example: 125 = 5 since 5 × 5 × 5 = 125. √ √ √ Another Helpful Rule - ab = a b - This rule is often used in simplifying radicals. Exponent Rules: Student should be familiar with the following rules and be able to apply them in problems. • (am )(an ) = am+n • (am ) ÷ (an ) = am−n , a 6= 0 • (am )n = amn • (ab)m = am bm • ( ab )m = an bn , b 6= 0 Exponents and Radicals: Students should also know the following relationships between exponents and radicals. • a−n = 1 an , a 6= 0 √ m • a n = ( n a)m , n > 0 Problems similar to the ones explored below will appear on the provincial exam. √ 1. Simplify 116 In this problem you are asked to simplify a radical that is not a perfect square. In other words there is no whole number x such that x × x = 116. When we are asked to simplify such a radical we are really being asked to examine the factors of 116, find the largest perfect square factor and remove it from the radical. I will start out by guessing that 4 is a factor of 116. On examining it I find that 116 = 4 × 29, now since 29 is prime (only factors are 1 and 29) my work is finished. 4 is a perfect square so I can rewrite my radical in the following form: 1 √ √ 116 = 4 × 29 √ √ √ 116 = 4 29 √ √ 116 = 2 29 √ √ 2 29 is the simplified form of 116 and the answer to this problem. √ 2. Simplify 3 −216 This problem is similar to problem 1 except instead of looking for perfect square factors of 216, we are looking for perfect cube factors. The negative sign may seem a bit daunting at √ 3 first but remember that −1 × −1 × −1 = −1 thus −1 = −1. Unlike square roots, cube roots are defined for negative numbers. To start off lets look for factors of 216 that are perfect cubes. First I’ll list some perfect cubes (remember these are just whole numbers multiplied by themselves 2 times). 1, 8, 27, 64, 125, 216, .... In my list above I notice that 216 = 6 × 6 × 6 and therefore −216 = −6 × −6 × −6 . Thus √ 3 −216 = −6 3. Which pattern could be used to predict 5−3 . 53 52 51 A. 50 5−1 5−2 5−3 125 25 5 1 53 52 51 C. 50 5−1 5−2 5−3 125 25 5 1 -5 -25 -125 1 5 1 25 1 125 53 52 51 B. 50 5−1 5−2 5−3 53 52 51 D. 50 5−1 5−2 5−3 Knowing the property that a−n = 1 know that 5−3 = 513 = 125 . 1 an , 15 10 5 0 − 15 1 − 10 1 − 15 15 10 5 0 -5 -25 -125 a 6= 0 can certainly simplify this problem. Thus we Upon examining the options above it is easy to conclude that B and D are unlikely to be correct. They do not even compute the values correctly. In deciding between option A and C we see that in A the negative exponent is dealt with according the the definition, therefore that is the pattern that can be used to predict 5−3 . 2 3 4. Evaluate 32 5 √ m For this problem refer to the definition a n = ( n a)m , n > 0 √ 3 32 5 = ( 5 32)2 √ We now need to consider 5 32 - Can we find a whole number that multiplied by itself 4 times is 32? In other words is there and √ x such that x × x × x × x × x = 32? I happen to know that 2 is such a number. Therefore: 5 32 = 2 √ 3 32 5 = ( 5 32)2 = 22 = 4 3 5. Evaluate 9− 2 In this problem we will use the same definition to first note that √ 3 9− 2 = ( 9)−3 Since the square root of 9 is easy to compute we now have: √ 3 9− 2 = ( 9)−3 = (2)−3 Since a−n = 1 an we have: √ 3 9− 2 = ( 9)−3 = (2)−3 = 3 1 23 = 1 8