Section 3: Division Properties of Radicals

Elementary and Intermediate Algebra: Graphs & Models (3rd Edition)

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Section 9.3 Division Properties of Radicals 907 Version: Fall 2007 9.3 Division Properties of Radicals Each of the equations x 2 = a and x 2 = b has a unique positive solution, x = a and x = b , respectively, provided a and b are positive real numbers. Further, because they are solutions, they can be substituted into the equations x 2 = a and x 2 = b to produce the results ( a ) 2 = a and ( b ) 2 = b, respectively. These results are dependent upon the fact that a and b are positive real numbers. Similarly, the equation x 2 = a b has the unique positive solution x = a b , provided a and b are positive real numbers. However, note that a b 2 = ( a ) 2 ( b ) 2 = a b , making a/ b a second positive solution of x 2 = a/b . However, because a/b is the unique positive solution of x 2 = a/b , this forces a b = a b . This discussion leads us to the following property of radicals. Property 1. Let a and b be positive real numbers. Then, a b = a b . This result can be used in two distinctly different ways. You can use the result to divide two square roots, as in 13 7 = 13 7 . Copyrighted material. See: http://msenux.redwoods.edu/IntAlgText/ 1
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908 Chapter 9 Radical Functions Version: Fall 2007 You can also use the result to take the square root of a fraction. Simply take the square root of both numerator and denominator, as in 13 7 = 13 7 . It is interesting to check these results on a calculator, as shown in Figure 1 . Figure 1. Checking that 13 / 7 = 13 / 7 . Simple Radical Form Continued David and Martha are again working on a homework problem. Martha obtains the solution 1 / 12 , but David’s solution 1 / (2 3) is seemingly different. Having learned their lesson in an earlier assignment, they use their calculators to find decimal ap- proximations of their solutions. Martha’s approximation is shown in Figure 2 (a) and David’s approximation is shown in Figure 2 (b). (a) Approximating Martha’s 1 / 12 . (b) Approximating David’s 1 / (2 3) . Figure 2. Comparing Martha’s 1 / 12 with David’s 1 / (2 3) . Martha finds that 1 / 12 0 . 2886751346 and David finds that 1 / (2 3) 0 . 2886751346 . They conclude that their answers match, but they want to know why such different looking answers are identical. The following calculation shows why Martha’s result is identical to David’s. First, use the division property of radicals ( Property 1 ) to take the square root of both numerator and denominator. 1 12 = 1 12 = 1 12
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Section 9.3 Division Properties of Radicals 909 Version: Fall 2007 Next, use the “first guideline for simple radical form” and factor a perfect square from the denominator. 1 12 = 1 4 3 = 1 2 3 This clearly demonstrates that David and Martha’s solutions are identical. Indeed, there are other possible forms for the solution of David and Martha’s home- work exercise. Start with Martha’s solution, then multiply both numerator and de- nominator of the fraction under the radical by 3.
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