Stitz-Zeager_College_Algebra_e-book

Solution 1 entering the data into the calculator

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Unformatted text preview: our chagrin, that the quadratic x2 − 2x − 1 doesn’t factor nicely. Hence, we resort to the quadratic formula to solve x2 − 2x − 1 = 0, and √ arrive at x = 1 ± 2. As before, these zeros divide the number line into three pieces. To √ √ help us decide on test values, we approximate 1 − 2 ≈ −0.4 and 1 + 2 ≈ 2.4. We choose x = −1, x = 0, and x = 3 as our test values and find f (−1) = 2, which is (+); f (0) = −1 which is (−); and f (3) = 2 which is (+) again. Our solution to x2 − 2x − 1 > 0 is where √ √ we have (+), so, in interval notation −∞, 1 − 2 ∪ 1 + 2, ∞ . To check the inequality x2 > 2x + 1 graphically, we set g (x) = x2 and h(x) = 2x + 1. We are looking for the x values where the graph of g is above the graph of h. As before we present the graphs on the right and the sign chart on the left. y 8 7 6 5 (+) (−) (+) 0 0 √ √ 1− 2 1+ 2 −1 0 3 4 3 2 −1 1 2 3 x 3. To solve 9x2 +4 ≤ 12x, as before, we solve 9x2 − 12x +4 ≤ 0. Setting f (x) = 9x2 − 12x +4 = 0, we...
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