Section 5: Radical Equations

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

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Section 9.5 Radical Equations 945 Version: Fall 2007 9.5 Radical Equations In this section we are going to solve equations that contain one or more radical ex- pressions. In the case where we can isolate the radical expression on one side of the equation , we can simply raise both sides of the equation to a power that will eliminate the radical expression. For example, if x 1 = 2 , (1) then we can square both sides of the equation, eliminating the radical. ( x 1 ) 2 = (2) 2 x 1 = 4 Now that the radical is eliminated, we can appeal to well understood techniques to solve the equation that remains. In this case, we need only add 1 to both sides of the equation to obtain x = 5 . This solution is easily checked. Substitute x = 5 in the original equation (1) . x 1 = 2 5 1 = 2 4 = 2 The last line is valid because the “positive square root of 4” is indeed 2. This seems pretty straight forward, but there are some subtleties. Let’s look at another example, one with an equation quite similar to equation (1) . Example 2. Solve the equation x 1 = 2 for x . If you carefully study the equation x 1 = 2 , (3) you might immediately detect a difficulty. The left-hand side of the equation calls for a “positive square root,” but the right-hand side of the equation is negative. Intuitively, there can be no solutions. A look at the graphs of each side of the equation also reveals the problem. The graphs of y = x 1 and y = 2 are shown in Figure 1 . Note that the graphs do not intersect, so the equation x 1 = 2 has no solution. However, note what happens when we square both sides of equation (3) . ( x 1) 2 = ( 2) 2 x 1 = 4 (4) Copyrighted material. See: http://msenux.redwoods.edu/IntAlgText/ 1
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946 Chapter 9 Radical Functions Version: Fall 2007 x y y = x 1 y = 2 Figure 1. The graphs of y = x 1 and y = 2 do not intersect. This result is identical to the result we got when we squared both sides of the equation x 1 = 2 above. If we continue, adding 1 to both sides of the equation, we get x = 5 . But this cannot be correct, as both intuition and the graphs in Figure 1 have shown that the equation x 1 = 2 has no solutions. Let’s check the solution x = 5 in the original equation (3) . x 1 = 2 5 1 = 2 4 = 2 Because the “positive square root of 4” does not equal 2, this last line is incorrect and the “solution” x = 5 does not check in the equation x 1 = 2. Because the only solution we found does not check, the equation has no solutions. The discussion in Example 2 dictates caution. Warning 5.
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Section 5: Radical Equations - Section 9.5 Radical...

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