Algebra_trig_sol

Algebra_trig_sol - Solutions to Algebra/Trig Review...

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Solutions to Algebra/Trig Review Problems for M408C 1. xx x 2 21 0 + Solution: First factor as much as possible to get x () + 1 0. The sign of the expression on the left can only change when either the numerator is 0 or the denominator is 0, i.e. at –0.5, 0, or 1. Make a sign chart and verify that the solutions lie in the intervals (] 05 0 ., and 1 , [) . 2. += 23 Solution: Substitute ux = to get the equation uu 2 . Solve this quadratic to get u =− 3 and u = 1 as two solutions. There will be no values of x satisfying x 3, so we only need to solve x = 1. This has only one solution: x = 1. 3. / / 11 0 22 2 1 2 12 −− = x Solution: Multiply through the equation by / 1 2 x to get rid of the negative exponent and the resulting equation is 10 = . So 2 x = and therefore the two solutions are x 1 2 . 4. || | | −+ += 123 4 Solution: For this problem, note that , , x −= −≥ −+ < 1 for for and , , 3 2 3 2 x +≥ < for for . We have to consider three separate cases: Case 1: x 1 Then | | x x x −+ +=−+ += + 1
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This note was uploaded on 11/30/2010 for the course M 408c taught by Professor Mcadam during the Spring '06 term at University of Texas.

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Algebra_trig_sol - Solutions to Algebra/Trig Review...

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