14 Questions with Answers - College Algebra - Examination 1 | MATH 121, Exams of Algebra

Download 14 Questions with Answers - College Algebra - Examination 1 | MATH 121 and more Exams Algebra in PDF only on Docsity! Math 121, Test 1, 23 January 2004 Name: ANSWERS Instructions. Do all of the following 14 problems. Each problem is worth 5pts. Show all appropriate details in your solutions. Calculators are not allowed. Good Luck! 1. Write the expression |x− 5|+ |2x + 2| given 0 < x < 4 without absolute value symbols. Answer. |x− 5|+ |2x + 2| = −(x− 5) + 2x + 2 = x + 7 if 0 < x < 4. 2. Simplify the complex fraction 1 x − 3 x+1 2 + 5 x+1 . Answer. Simplify by multiplying the numerator and denominator by the LCD x(x + 1): 1 x − 3 x+1 2 + 5 x+1 = ( 1 x − 3 x+1 ) x(x + 1)( 2 + 5 x+1 ) x(x + 1) = x + 1− 3x 2x(x + 1) + 5x = 1− 2x x(2x + 7) . 3. Factor the expression 2x + 1− 4xy − 2y. Answer. 2x + 1− 4xy − 2y = 1(2x + 1)− 2y(2x + 1) = (1− 2y)(2x + 1). 4. Simplify: (2a−1bc−2)2 (3−1b)(2−1ac−2)3 . Answer. (2a−1bc−2)2 (3−1b)(2−1ac−2)3 = 4a−2b2c−4 3−12−3a3bc−6 = 96bc2 a5 . 5. Simplify: x− 2 x2 − 6x + 9 ÷ x 2 − 4 x2 − x− 6 . Answer. x− 2 x2 − 6x + 9 ÷ x 2 − 4 x2 − x− 6 = x− 2 (x− 3)(x− 3) · (x− 3)(x + 2) (x + 2)(x− 2) = 1 x− 3 6. Write the complex number 4− 3i 3 + 2i in standard form. Answer. 4− 3i 3 + 2i = (4− 3i)(3− 2i) (3 + 2i)(3− 2i) = 6− 17i 13 = 6 13 − 17 13 i 7. Solve the equation 3x + 7 2x− 3 = −2. 1 Answer. Multiply both sides of the equation by (2x− 3) to obtain: 3x + 7 = −2(2x− 3), or 3x + 7 = −4x + 6. Therefore, 7x = −1, and so x = −1 7 . 8. Pump 1 can drain a pool in 6 hours, while pump 2 can drain the same pool in 7 hours. How long would it take to drain the pool using both pumps? Answer. Let t be the amount of time in hours it would take both pumps to drain the pool. Then t 7 + t 6 = 1 and so multiplying both sides of this equation by 42 yields 6t + 7t = 42. Therefore, t = 42 13 hours, or 3 3 13 hours. 9. Solve the quadratic equation 1 3 x2 + 1 6 x + 2 = 0 by using the quadratic formula. Answer. Multiply both sides of this equation by 6 to eliminate the fractions. Therefore, we solve 2x2 + x + 12 = 0 using a = 2, b = 1 and c = 12 in the quadratic equation. This leads to x = −1± √ 1− 4(2)(12) 4 = −1 4 ± i √ 95 4 10. Solve the radical equation √ 3x + 7− 1 = x. Check all proposed solutions. Answer. Rewrite the equation as √ 3x + 7 = x + 1 and then square both sides to obtain 3x + 7 = x2 + 2x + 1, or x2 − x − 6 = 0. Therefore, (x − 3)(x + 2) = 0, and so x = 3 and x = −2 are proposed solutions. Plugging these back into the original equation, we find that x = 3 works, while x = −2 does not. Therefore, the solution is x = 3. 11. Solve the equation (2x + 1) 2 3 + (2x + 1) 1 3 = 6. Answer. Let u = (2x + 1) 1 3 to convert this to the equation u2 + u − 6 = 0. Therefore, (u + 3)(x− 2) = 0 and so u = −3 or u = 2. Therefore, 2x + 1 = −27 or 2x + 1 = 8. This leads to solutions x = −14 and x = 7/2. 12. Solve the inequality |3x− 6| > 15. Answer. The absolute value inequality |3x−6| > 15 implies that 3x−6 > 15 or 3x−6 < −15. Solving these individually we find that x > 7 or x < −3. 13. Solve the inequality x + 2 x− 5 ≤ 2 and write the answer in interval notation. 2