Math 121, Chapter 1 and Complex Numbers Practice Questions Hints and Answers. 2. Multiply numerator and denominator by complex conjugate and simplify:

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1 Math 121, Chapter 1 and Complex Numbers Practice Questions Hints and Answers 1. (a) (3 2i)( + 1i) = i 8i 2i 2 = 12 5i 2( 1) = 1 5i. (b) i 223 = (i ) 55 (i 3 ) = i. 2. Multiply numerator and denominator by complex conjugate and simplify: (3 2i)( 5i) ( + 5i)( 5i) = 12 23i + 10i2 1 = 2 23i 1 = i. 3. (a) 2x + 5 3x 1 = 1 implies 2x + 5 = 3x 1 implies 6 = x. Thus, the solution is x = 6. (b) 3 x d = c where c > 0 implies x d = c/3 and so x d = c/3 or x d = c/3. Therefore, x = d + c/3 or x = d c/3. (c) Multiply both sides by 15 to obtain 5(21 2x) = 3(x + 6) and so x = 12x + 18 or 22x = 87 and so x = 87/22. (You should plug this back in to check your answer). (d) Distributing we have 10s 22 8s 36 = 0 and so 2s 58 = 0 or 2s = 58 and so s = 29. (As usual, you should plug this back in original equation to check.) (e) We will isolate the expression x + 7 first. Thus 3 x + 7 = 33 and so x + 7 = 11. Then x + 7 = 11 or x + 7 = 11 Thus x = or x = 18. The solutions are 18,. (f) Solve t = 70. This implies t = 70 or t = 70. Then 60 = 30t or 200 = 30t. Thus t = 2 or t = 20/ Thus Rueben will be exactly 70 miles from Barstow in 2 hours and in 6.7 hours.. (a) Let A be the amount of time in hours that it would take the assistant to build the fence alone. Because the two workers can finish the fence in 5 hours time, we get that Therefore, 5 A = 3 8 and so A = 0 3 hours A = 1. (b) The rate for the painter is 1/1 and the rate for the assistant is 1/35. Use the formula that rate times time is the portion of the job completed. Let t be the time needed for the two working together to paint the kitchen. Then t t 1 35 = 1 or 5t + 2t = 70 and so t = 10. It would take 10 hours for the two working together to paint the kitchen. 1

2 5. Let x = the amount of antifreeze solution that should be drained and replaced with pure antifreeze. Then there are 6 x liters of 20% solution, and x liters of 100% solution to produce 6 liters of 50% solution. The equation for the amount of antifreeze in the radiator is thus: (6 x)(.2) + x(1.0) = (.5)(6). Therefore, 1.2.2x + x = 3 and so x = 3 and so.8x = 1.8 and so x = 9 liters. 6. (a) A = 1h(b b 2 ) implies b 1 + b 2 = 2A implies b h 2 = 2A b h 1 implies b 2 = 2A b 1h. h (b) Multiplying both sides of 1 R = 1 R R 2 by R R 1 R 2 implies R 1 R 2 = RR 2 +RR 1. Therefore, R 1 R 2 RR 2 = RR 1 and so R 2 (R 1 R) = RR 1, and finally, R 2 = RR 1 R 1 R. (c) Solve similarly to (b), or 1 R = 1 R R 2 implies R = multiplying numerator and denominator by R 1 R 2 to obtain R = R 1R 2 R 2 + R R Simply this fraction by R 2 7. (a) Let x be the amount invested at 8%. Then interest earned = interest rate times amount invested yields the equation (2500)(.055) +.08x =.07( x) which has solution x = Thus $3750 is invested at 8%. (b) Let x be the amount invested at 11%. Then x is the amount invested at 8%. Therefore, the interest earned equation is 0.08(11000 x) x = 108 This simplifies to.03x = (11000) = 168, and so x = 168/.03 = 5600, which means x = 500. Thus $500 was invested at 8% and $5600 was invested at 11%. 8. (a) Let x be the number of grams of pure gold added. Then computing the amount of gold in the allow yields ( ) x = 18 (x + 15) 2 2 which has solution x = 10 grams. (b) Let x be the number of $6.50 per ounce silver needed in the mixture. Then 0 x is the number of $9.00 per ounce silver that is needed. This leads to the equation 6.5x + 9(0 x) = 0(8). Then, 2.5x = 320, or 5x/2 = 0, and so x = 16. Therefore, 16 ounces of $6.50 per ounce silver should be mixed with 2 ounces of $9.00 per ounce silver. 2

3 9.(a) Let d be the distance each direction, and let t be the amount of time it took the messenger to run to the palace. Using d = rt we obtain the equation 10t = (8.75 t). Therefore, 1t = 35 and so t = 2.5 hours. Therefore, the distance from the messenger s residence to the palace is d = 10 miles 2.5hours = 25 miles. hours (b) Let t be the time in seconds it took the sprinter to run to the end of the track. Then the return trip took 126 t seconds. Then t = length of track = 2(126 t). Therefore, t = 2(126 t) or 6t = 2(126) or t = 2 seconds. Then the track is (2) = 168 meters long x 2 + 3x = implies 2x 2 + 3x = 0. Now the quadratic equation implies x = 3 ± 3 2 (2)( ), thus the solutions are x = 3 ± 1. 2(2) 11. Using the difference of two squares pattern, we factor (x 5) 2 9 = ((x 5) + 3)((x 5) 3) Therefore, (x 2)(x 8) = 0 and so x = 2 or x = First, 1 2 x2 + 3 x + 1 = 0 implies that (multiply both sides of the equation by ) 2x 2 + 3x + = 0. Now, for the purposes of the quadratic equation, a = 2, b = 3 and c =, and so x = 3 ± 3 2 (2)() 2(2) = 3 ± 23. Therefore, the answer is x = 3 ± i 23 = 3 ± i (a) Let w be the width, and l be the length. The amount of fence used will be 2l + 3w (the 3w is because of the extra fence in the middle to create the two smaller rectangles). Now 2l + 3w = 600 and so l = (600 3w)/2. Also, A = lw and so we get that w(300 3 w) = 15, Thus, 3 2 w2 300w + 15, 000 = 0. Multiplying both sides by 2 to get rid of the faction, we have 3w 2 600w + 30, 000 = 0. This factors as 3(w 2 200w + 10, 000) = 3(w 100) 2. Therefore, 3(w 100) 2 = 0 and so w = 100. Hence l = (600 3(100))/2 = 150. So the enclosure s dimensions are 100 feet by 150 feet. 3

4 (b) The ball will have a height of 130 feet when 16t t + 7 = 130, or 16t 2 88t = 0. There are no real solutions to this quadratic equation (check with the quadratic formula), so the ball never reaches a height of 130 feet. The ball will land when h(t) = 0. Thus one should solve 16t t + 7 = 0. This has solutions t = 88 ± 88 2 ( 16)(7) 2( 16) = 88 ± so t is approximately 5.58 seconds or.078 seconds. But the equation is only for t 0, (time after it was thrown), so the ball hits the ground after approximately 5.58 seconds. (c) The volume is 3(x 6) 2 and so we solve 3(x 6) 2 = Then (x 6) 2 = and so x 6 = and thus x 6 = 8.5, or x = 1.5. Thus a sheet of cardboard that measures 1.5 inches by 1.5 inches is needed. (d) Let the screen have dimensions x by y. Then y = and so y = x. By the Pythagorean x 3 3 theorem ( ) = x 2 + y 2 = x x = x x2 (9)(5) and so (9)(5) 2 = 9x x 2 2 and then x = The screen measures 32. inches by 3.2 inches. = (3)(5) 5 = 32. and y = 3 (32.) = 1. (a) Rewrite as 2x = x + 1 and then square both sides to obtain (2x 5) + 2 2x = x + 1 and simplify to x 5 = 2 2x 5 and square both sides again to obtain x 2 10x + 25 = (2x 5). Therefore, x 2 18x + 5 = 0, and so (x 3)(x 15) = 0. Consequently, the proposed solutions are x = 3 or x = 15. However, x = 15 doesn t work, therefore x = 3 is the only solution. (b) Rewrite the equation as 3x + 7 = x + 1 and then square both sides to obtain 3x + 7 = x 2 + 2x + 1, or x 2 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 = Raise both sides of the equation to the th power to get that 256x 3 = x 2 Therefore, x 2 (256x 1) = 0. Thus the proposed solutions to the equation are x = 0 and x = Plugging them back into the equation, we can see that they both are solutions.

5 16. (a) Make the substitution u = (3x 5) 1 3 to convert this into the quadratic equation u 2 + 6u + 8 = 0. Therefore, (u + )(u + 2) = 0 and so u = 2 or u =. Now u 3 = 3x 5 and so x = u Consequently, the solutions to the equation are x = = 1 and x = = (b) x 5x 2 + = 0 implies (x 2 )(x 2 1) = 0 and so x = 2, 2, 1, 1. (c) x 6 5x = 0 implies (x 3 1)(x 3 + 6) = 0 and so x = 1 and x = 3 6 are the real solutions. 17. Pump 1 can drain t 10 of the pool in t hours and pump 2 can drain t of the pool in t hours. 8 So, let x be the number of hours it takes the pumps to drain the pool together. Then x 10 + x 8 = 1. Multiplying both sides of the equation by 80, we get 8x + 10x = 80. Therefore, 18x = 80 and so x = 80/18 = 0/9. Thus, together the pumps can drain the pool in 9 hours. 18.(a) First, x 3 6x 2 + 8x = x(x 2)(x ). Therefore, x(x 2)(x ) = 0. Thus, the solutions are x = 0, x = 2 and x =. (b) Factor by grouping: x 2 (5x + 6) 16(5x + 6) = 0, or (x 2 16)(5x + 6) = 0. Therefore, x =, x = or x = 6/5 19. Write the equation as x 1 = 2x 8 so that the radical is alone on one side of the equation. Squaring both sides of this equation, we get x 1 = x 2 32x + 6, bringing everything to one side yields x 2 36x + 65 = 0. Factoring this quadratic, one obtains (2x 5)(2x 13) = 0. Thus the proposed solutions are x = 5 13 and x = 2 2. Plugging x = 5 back into the original 2 equation leads to 3 = 3, hence 5 13 is not a solution. Plunging x = into the original equation 2 2 leads to 5 = 5, and so it is a solution. Therefore, the original equation has solution x =

6 20. (a) 1 5x 6 < 9 is equivalent to 5 5x < 15 (by adding 6 to all sides of the inequality), which in turn is equivalent to 1 x > 3 (dividing by 3 on all sides). Therefore, the solution, in interval notation, is ( 3, 1]. (b) Let x represent the percentage on the final test. Then 75 (.7)(92) + (.3)x 85, i.e., x 85. Subtracting 6. from each side yields x Dividing each side by.3 leads to x Thus the student needs between 35.33% and 68.67% on the final exam. 21. x 2 10 is equivalent to x 2 10 or x Solving these individually leads to x 2 or x 3. In interval notation, the answer is: (, 2] [3, ). 22. (, ) because an absolute value is always larger than a negative number. 23., the emptyset, because an absolute value is never negative. 2. 6x + 3 < 15 is equivalent to 15 < 6x + 3 < 15 which leads to 3 < x < 2, or in interval notation: ( 3, 2). 25. x 2 < x + 30 is equivalent to x 2 + x 30 < 0, or (x + 6)(x 5) < 0. The critical values are 5 and 6. Next use a sign chart: x < 6 x = 6 6 < x < 5 x = 5 5 < x x + 6 neg 0 pos pos pos x 5 neg neg neg 0 pos (x + 6)(x 5) pos 0 neg 0 pos The inequality is true when the product is negative, thus the solution in interval notation is found to be ( 6, 5). 26. The inequality x + 1 x 2 yields x + 5 x 2 using this information. x is equivalent to 2 0. Simplifying the left hand side, x 2 0. Thus the critical values are x = 2 and x = 5. Next create a sign chart x < 2 x = 2 2 < x < 5 x = 5 5 < x x 2 neg 0 pos pos pos x + 5 pos pos pos 0 neg x + 5 x 2 neg und pos 0 neg In interval notation the solution is (2, 5]. 6

7 27. The problem suggest v 0 = 80 and s 0 = 33. Thus s = 16t t We then solve the inequality 16t t + 33 > 97. Thus 16t t 6 > 0 and multiplying both sides by 1 we have 16t 2 80t + 6 < 0 and dividing by 16 yields t 2 5t + < 0 and factoring gives us (t )(t 1) < 0. The critical values are 1 and. t < 1 t = 1 1 < t < t = < t t 1 neg 0 pos pos pos t neg neg neg 0 pos (t 1)(t ) pos 0 neg 0 pos Thus the inequality is true when 1 < t <. That is the ball is more than 97 feet above the ground between 1 second and seconds after it was thrown. 28. We look for when the expression is positive or 0 (note it is undefined at x = and it is 0 when x = 2 or x = 1), so we use the sign chart: x < 1 x = 1 1 < x < 2 x = 2 2 < x < x = < x x + 1 neg 0 pos pos pos pos pos x 2 neg neg neg 0 pos pos pos x neg neg neg neg neg 0 pos (x 2)(x + 1) x neg 0 pos 0 neg und pos Thus the inequality is true when 1 x 2 or when x >. In interval notation, the answer is [ 1, 2] (, ) 29. The combined variation equation is L = kd. To find k we plug in the given values: h2 6 = k2. Therefore, k = 600/16 = Now, if a column has a diameter of 3 feet and a 102 height of 1 feet, it can support up to L = (37.5)(3 ) tons. 30. Increases by a factor of 2, i.e., can support a weight 16 times greater. 7

8 31. Under these parameters, the modified column could support (.5) th of the weight of the 32 original column, that is, it can support only 1 of the amount of weight (a) Use the relation I = k where I is the sound intensity,and d is the distance from d2 the engine to the ramp meter. For the given data, k = (0.5)(25) = Thus the intensity when d = 11 is I = Thus to three decimal places, the sound intensity is W/m 2. (b) (i) If the distance doubles, the sound intenisty is one-quarter of the original intensity. (ii) If the distance triples, the sound intensity will be one-ninth of the original intensity. We obtain this from I = k, and so replacing the distance with 3d we obtain d2 I = which is 1/9 of the original intensity. k (3d) 2 = k 9d = k d The relation is L = kwd2 l new beam can support. We use 600 = k(6)(82 ) 1 L = 700(5.5)(62 ) 32(16) to find k = (1)(600) (6)(6) Thus the new beam can support approximately 271 pounds. = 700. Thus the 32 8