Stitz-Zeager_College_Algebra_e-book

# A y 70 40 0 1 20 3 2 1 12 ha y 0 y f t et 3 t 2

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Unformatted text preview: eed 2 − 4 x + 3 ≥ 0. √ While it may be tempting to write this as 2 ≥ 4 x + 3 and take both sides to the fourth power, there are times when this technique will produce erroneous results.6 Instead, we solve √ √ 2 − 4 x + 3 ≥ 0 using a sign diagram. If we let r(x) = 2 − 4 x + 3, we know x ≥ −3, so we concern ourselves with √ only this portion of the number line. To √ ﬁnd the zeros of r we set √ 4 r(x) = 0 and solve 2 − 4 x + 3 = 0. We get 4 x + 3 = 2 so that 4 x + 3 = 24 from which we obtain x + 3 = 16 or x = 13. Since we raised both sides of an equation to an even power, we need to check to see if x = 13 is an extraneous solution.7 We ﬁnd x = 13 does check since √ √ √ 2 − 4 x + 3 = 2 − 4 13 + 3 = 2 − 4 16 = 2 − 2 = 0. Below is our sign diagram for r. (+) 0 (−) −3 13 √ We ﬁnd 2 − 4 x + 3 ≥ 0 on [−3, 13] so this is the domain of g . To ﬁnd a sign diagram for g , √ we look for the zeros of g . Setting g (x) = 0 is equivalent to 2 −...
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## This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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