Tiêu đề MATH 20 LE 1 ANSWER KEY.pdf ✅

Math 20 Sample 1st Long Exam First Semester, A.Y. 2025-2026 MBAN 109 This sample long exam and accompanying answer key are intended as reviewers for the actual long exam. General Instructions: This exam is good for 80 minutes. Provide neat and complete solutions. Calculators are not allowed. 1. Find the solution sets of the following equations. (a) x + 1 2x 2 − x − 6 + x + 5 2x 2 − 11x − 21 = x − 1 x 2 − 9x + 14 − 1 2x + 3 (b) 3 √3 x + 2 − 1 + √3 x + 2 − 1 2 = 0 (c) |2x − 3| = 5 Solution: (a) In solving the equation, it is recommended to express the denominators as a product of factors: x + 1 (2x + 3)(x − 2) + x + 5 (2x + 3)(x − 7) = x − 1 (x − 2)(x − 7) − 1 2x + 3 . Now, the LCD can easily be calculated by analyzing the factors of the denominator. The LCD in this case is (x − 2)(x − 7)(2x + 3). There is no need to simplify the LCD. Multiplying the LCD to both sides gives the equation (x + 1)(x − 7) + (x + 5)(x − 2) = (x − 1)(2x + 3) − (x − 2)(x − 7). In this case, it is necessary to simplify the factors in order to combine like terms. Thus, (x 2 − 6x − 7) + (x 2 + 3x − 10) = (2x 2 + x − 3) − (x 2 − 9x + 14) 2x 2 − 3x − 17 = x 2 + 10x − 17 x 2 − 13x = 0 x(x − 13) = 0 =⇒ x = 0 or 13. One may verify that these values indeed satisfy the original equation, so we have no extra- neous solutions. ∴ The solution set of the equation is {0, 13}. (b) Let u = √3 x + 2 − 1. Then, the equation can be written as a quadratic form in u. 3u + u 2 = 0 u(3 + u) = 0 Thus, u = 0 or u = −3. We now examine each case. if u = 0, if u = −3, √3 x + 2 − 1 = 0 √3 x + 2 − 1 = −3 √3 x + 2 = 1 √3 x + 2 = −2 x + 2 = 13 x + 2 = (−2)3 x = −1 x = −10. One may verify that these values indeed satisfy the original equation, so we have no extra- neous solutions. ∴ The solution set of the equation is {−1, −10}.