Survey
* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
Chapter 15 Radical Expressions and Equations Notes 15.1 Introduction to Radical Expressions The symbol is called the square root and is defined as follows: c2 = a a = c only if Sample Problem: Solution: Simplify 16 = 4 16 since 4 2 = 16 . Note that every positive number has two square roots, a positive and a negative root. For example, the square roots of 16 are 4 and -4, since 4 2 = 16 and (−4) 2 = 16 . The symbol implies the positive root, or the principal square root. To get the negative root, a negative sign must be used in front of the square root sign as in − Sample Problem: Simplify Solution: − 16 = −4 − 16 Student Practice: Simplify each of the following radical expressions. 1. 25 2. − 81 3. 1 4. − 36 5. 100 6. − 49 7. − 9 (not real #.) 8. y2 . 9. x6 11. x 16 13. 9 x 12 10. w10 12. 16a 2 b 8 Tip: When taking roots of exponential expressions, keep the base and take half of the exponent. 15.2 Simplifying Radical Expressions ● SIMPLIFYING RADICAL EXPRESSIONS A radical expression is simplified when there are no perfect square factors inside the radical; i.e., when you take as much as you can out of the radical. Simplifying can be done using the following rule: PRODUCT RULE a•b = a • b a • b = a•b OR The product rule can be used to simplify radicands that are not perfect squares. Simply factor the radicand using a perfect square as a factor. Sample Problem 1: Simplify 36 36 = 4 • 9 = 4 • 9 = 2 • 3 = 6 Solution: Sample Problem 2: Simplify 45 Solution: Since 45 is not a perfect square, we first factor 45 using a perfect square factor. It may help to list the first few perfect square factors, which are 1,4,9,16: In our 45 = 9 • 5 = 9 • 5 = 3 • 5 = 3 5 case, we’ll use 9. ______________________________________________________________________ FOR LARGER NUMBERS. For larger numbers, it can be helpful to factor the radicand into prime factors. To do so, divide by the smallest prime number as man times as possible, then the next smallest, then the next until you are left with a prime number. For each pair of identical factors, . 5 Sample Problem 2: Simplify 240 3 15 2 2 2 2 30 60 120 240 Solution: 240 = 2 • 2 • 2 • 2 • 3 • 5 = 4 • 4 • 3 • 5 = 4 • 4 • 3 • 5 = 2 • 2 • 15 = 4 15 2 2 240 = 2 • 2 • 15 = 4 15 Tip: To check your answer, multiply the outer number twice and then multiply by the inner number to get what you started with. 4 • 4 • 5 = 240 Sample Problem 3: Solution: Simplify x 13 x 13 = x 12 • x = x 12 • x = x 6 x Student Practice: Simplify each radical expression. 1. 18 2. 50 3. 27 4. 24 5. 32 6. 72 8. 64 28. 21 9. w 8 10. x9 11. a 15 12. a3 13. a 5 b10 14. 20 x 7 15.3 Addition and Subtraction of Radical Expressions To add/subtract two radical terms, they MUST BE LIKE TERMS. In other words, the radicands MUST be the same. If the radicands are not the same, simplify the radicands to make them match and then combine the coefficients and leave the radicand alone. 4 72 + 7 8 Sample Problem: Add Solution: 4 72 + 7 8 = 4 36 • 2 + 7 4 • 2 = 4•6 2 + 7•2 2 = 24 2 + 14 2 = 38 2 Student Practice: Add or subtract as indicated. 1. 5 2 + 8 2 2. 2 10 x − 8 10 x 3. − 4 5 + 5 5 4. 54 + 2 24 5. 2 10 − 7 40 7. 5 2 x 2 − 2 200 6. 5 18 − 2 50 + 6 2 15.4 Multiplying and Dividing Radical Expressions When multiplying two radical expressions, recall the product rule from before which states that the product of two radical expressions is the radical of the product. PRODUCT RULE a • b = a•b Sample Problem: 12 x 3 y • 6 x 5 y 4 Multiply, then simplify Solution: 12 x 3 y • 6 x 5 y 4 = 12 x 3 y • 6 x 5 y 4 = 72 x 8 y 5 = 36 • 2 • x 8 • y 4 • y = 6 x 4 y 2 2 y Student Practice: Use the product rule to multiply, then simplify. 1. 50 • 2 2. 7• 7 3. 13 • 13 4. 3• 6 5. 2x 3 • 8x 3 y 4 6. x3 • x2 7. 3 5 • 2 10 To multiply radical expressions with more than one term, use the product rule discussed earlier along with the distributive property. Multiply the inside of the radicals together and the outside of the radicals together, then simplify if possible. Sample Problem: Multiply (2 3 − 5 2 )(3 3 + 2 ) Solution: (FOIL) ( 2 3 − 5 2 )(3 3 + 2 ) = 2 3 • 3 3 + 2 3 • 2 − 5 2 • 3 3 − 5 2 • 2 = 6 9 + 2 6 − 15 6 − 5 4 = 6 • 3 − 13 6 − 5 • 2 = 18 − 13 6 − 10 = 8 − 13 6 Student Practice: 1. 3 ( 5 + 2 ) Multiply and simplify. 3. (3 2 − a )(3 2 + a ) 2. (1 − 7 )(4 + 3 7) 4. (2 5 − 3 2 )( 5 + 4 2 ) 5. (5 + x ) 2 6. (a − b ) 2 • Dividing Radical Expressions When dividing rational expressions, use the quotient rule mentioned before stating that the quotient of two radicals is the radical of the quotient. QUOTIENT RULE a b Sample Problem: Divide and simplify. Solution: 24 x 11 3x 7 = = a b 24 x11 3x 7 24 x 11 = 8x 4 = 4 • 2 x 4 = 2 x 2 2 7 3x Student Practice: Divide and simplify. 7. 75 3 8. 48 y 9 3y3 9. 24 x 7 10. 3x 2 18 x 5 3x ● RATIONALIZING DENOMINATORS Often times in mathematics it is useful to write a fraction without a radical in the denominator. The process of writing a fraction with a radical in the denominator as an equivalent fraction without a radical in the denominator is called rationalizing the denominator. To rationalize a denominator, try the following: • Multiply the numerator and denominator by a radical term that will make the bottom radicand a perfect square. Sample Problem: Solution: 3 Rationalize 3 5 • 5 5 = 5 3 5 25 = 3 5 5 Student Practice: Rationalize each denominator. 10 11. 12. 3 3 7 13. 15. 5 14. x 4 3 2 16. 3 8 2 3 15.5 Solving Radical Equations To solve equations with radicals, 1. Isolate the radical on one side of the equation. 2. Raise each side to the power of the index of the radical. If the equation still contains a radical, repeats steps 1 and 2. 3. Solve the resulting equation. 4. Ensure that your answer works in the original equation. Sample Problem: Solve for x a. x−5 +2 = 7 x−5 +2 = 7 Solution: a. x−5 = 5 x − 5 = 25 x = 30 The solution is 30. Solution: b. 2x + 1 + 1 = x 2x + 1 = x − 1 2 x + 1 = ( x − 1) 2 2x + 1 = x 2 − 2x + 1 0 = x 2 − 4x 0 = x ( x − 4) x=0 x=4 Only x = 4 is a solution to the original equation. b. 2x + 1 + 1 = x 30 − 5 + 2 = 7 Check: 25 + 2 = 7 5+2 = 7 Student Practice: Solve each equation for x. 1. x + 11 = 15 2. 10 x − 1 − 6 = 1 3. 9 x + 10 + 5 = 15 4. 5x + 4 = x + 8 5. 24 + 2 x = x 7. 2 x + 1 + 6 = 3 6. 8 x + 32 − 4 = x