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 2, Chapter and Complex Numbers Practice Questions Hints and Answers (a) (3 2i)( + i) = 2 + 3i 8i 2i 2 = 2 5i 2( ) = 5i (b) i 223 = (i ) 55 (i 3 ) = i 2 Multiply numerator and denominator by complex conjugate and simplify: (3 2i)( 5i) ( + 5i)( 5i) = 2 23i + 0i2 = 2 23i = 2 23 i 3 (a) 2x + 5 3x = implies 2x + 5 = 3x 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 5 to obtain 5(2 2x) = 3(x + 6) and so 05 0x = 2x + 8 or 22x = 87 and so x = 87/22 (You should plug this back in to check your answer) (d) Distributing we have 0s 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 = Then x + 7 = or x + 7 = Thus x = or x = 8 The solutions are 8, (f) Solve 30 30t = 70 This implies 30 30t = 70 or 30 30t = 70 Then 60 = 30t or 200 = 30t Thus t = 2 or t = 20/3 67 Thus Rueben will be exactly 70 miles from Barstow in 2 hours and in 67 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 = (b) The rate for the painter is / and the rate for the assistant is /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 35 = or 5t + 2t = 70 and so t = 0 It would take 0 hours for the two working together to paint the kitchen

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 00% solution to produce 6 liters of 50% solution The equation for the amount of antifreeze in the radiator is thus: (6 x)(2) + x(0) = (5)(6) Therefore, 2 2x + x = 3 and so 2 + 8x = 3 and so 8x = 8 and so x = 9 liters 6 (a) A = h(b 2 + b 2 ) implies b + b 2 = 2A implies b h 2 = 2A b h implies b 2 = 2A b h h (b) Multiplying both sides of R = R + R 2 by R R R 2 implies R R 2 = RR 2 +RR Therefore, R R 2 RR 2 = RR and so R 2 (R R) = RR, and finally, R 2 = RR R R (c) Solve similarly to (b), or R = R + R 2 implies R = multiplying numerator and denominator by R R 2 to obtain R = R R 2 R 2 + R R + Simply this fraction by R 2 7 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 = 3750 Thus $3750 is invested at 8% 8 (a) Let x be the number of grams of pure gold added Then computing the amount of gold in the allow yields ( ) 5 + x = 8 (x + 5) 2 2 which has solution x = 0 grams (b) Let x be the number of $650 per ounce silver needed in the mixture Then 0 x is the number of $900 per ounce silver that is needed This leads to the equation 65x + 9(0 x) = 0(8) Then, 25x = 320, or 5x/2 = 0, and so x = 6 Therefore, 6 ounces of $650 per ounce silver should be mixed with 2 ounces of $900 per ounce silver 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 0t = (875 t) Therefore, t = 35 and so t = 25 hours Therefore, the distance from the messenger s residence to the palace is d = 0 miles 25hours = 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 26 t seconds Then t = length of track = 2(26 t) Therefore, t = 2(26 t) or 6t = 2(26) or t = 2 seconds Then the track is (2) = 68 meters long 2

3 0 2x 2 + 3x = implies 2x 2 + 3x = 0 Now the quadratic equation implies x = 3 ± 3 2 (2)( ), thus the solutions are x = 3 ± 2(2) Using the difference of two squares pattern, we factor (x 5) 2 9 = ((x 5) + 3)((x 5) 3) Therefore, ()(x 8) = 0 and so x = 2 or x = 8 2 First, 2 x2 + 3 x + = 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 23 3 (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) = 5, Thus, 3 2 w2 300w + 5, 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 + 0, 000) = 3(w 00) 2 Therefore, 3(w 00) 2 = 0 and so w = 00 Hence l = (600 3(00))/2 = 50 So the enclosure s dimensions are 00 feet by 50 feet (b) The ball will have a height of 30 feet when 6t t + 7 = 30, or 6t 2 88t + 23 = 0 There are no real solutions to this quadratic equation (check with the quadratic formula), so the ball never reaches a height of 30 feet The ball will land when h(t) = 0 Thus one should solve 6t t + 7 = 0 This has solutions t = 88 ± 88 2 ( 6)(7) 2( 6) = 88 ± so t is approximately 558 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 558 seconds (c) The volume is 3(x 6) 2 and so we solve 3(x 6) 2 = 2675 Then (x 6) 2 = 7225 and so x 6 = 7225 and thus x 6 = 85, or x = 5 Thus a sheet of cardboard that measures 5 inches by 5 inches is needed 3

4 (a) Rewrite as 2x 5 + = x + and then square both sides to obtain (2x 5) + 2 2x 5 + = x + and simplify to x 5 = 2 2x 5 and square both sides again to obtain x 2 0x + 25 = (2x 5) Therefore, x 2 8x + 5 = 0, and so (x 3)(x 5) = 0 Consequently, the proposed solutions are x = 3 or x = 5 However, x = 5 doesn t work, therefore x = 3 is the only solution (b) Rewrite the equation as 3x + 7 = x + and then square both sides to obtain 3x + 7 = x 2 + 2x +, 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 = 3 5 Raise both sides of the equation to the th power to get that 256x 3 = x 2 Therefore, x 2 (256x ) = 0 Thus the proposed solutions to the equation are x = 0 and x = 256 Plugging them back into the equation, we can see that they both are solutions 6 (a) Make the substitution u = (3x 5) 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 = u3 + 5 Consequently, the solutions to the equation are x = = and x = = (b) x 5x 2 + = 0 implies (x 2 )(x 2 ) = 0 and so x = 2, 2,, (c) x 6 5x = 0 implies (x 3 )(x 3 + 6) = 0 and so x = and x = 3 6 are the real solutions 7 Pump can drain t 0 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 0 + x 8 = Multiplying both sides of the equation by 80, we get 8x + 0x = 80 Therefore, 8x = 80 and so x = 80/8 = 0/9 Thus, together the pumps can drain the pool in 9 hours 8(a) First, x 3 6x 2 + 8x = x()(x ) Therefore, x()(x ) = 0 Thus, the solutions are x = 0, x = 2 and x = (b) Factor by grouping: x 2 (5x + 6) 6(5x + 6) = 0, or (x 2 6)(5x + 6) = 0 Therefore, x =, x = or x = 6/5

5 9 Write the equation as x = 2x 8 so that the radical is alone on one side of the equation Squaring both sides of this equation, we get x = x 2 32x + 6, bringing everything to one side yields x 2 36x + 65 = 0 Factoring this quadratic, one obtains (2x 5)(2x 3) = 0 Thus the proposed solutions are x = 5 3 and x = 2 2 Plugging x = 5 back into the original 2 equation leads to 3 = 3, hence 5 3 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 = (a) 5x 6 < 9 is equivalent to 5 5x < 5 (by adding 6 to all sides of the inequality), which in turn is equivalent to x > 3 (dividing by 3 on all sides) Therefore, the solution, in interval notation, is ( 3, ] (b) Let x represent the percentage on the final test Then 75 (7)(92) + (3)x 85, ie, x 85 Subtracting 6 from each side yields 06 3x 206 Dividing each side by 3 leads to 3533 x 6867 Thus the student needs between 3533% and 6867% on the final exam 2 0 is equivalent to 0 or 0 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 < 5 is equivalent to 5 < 6x + 3 < 5 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) 5

6 26 The inequality x + yields x + 5 using this information x + 2 is equivalent to 2 0 Simplifying the left hand side, 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 neg 0 pos pos pos x + 5 pos pos pos 0 neg x + 5 neg und pos 0 neg In interval notation the solution is (2, 5] 27 The problem suggest v 0 = 80 and s 0 = 33 Thus s = 6t t + 33 We then solve the inequality 6t t + 33 > 97 Thus 6t t 6 > 0 and multiplying both sides by we have 6t 2 80t + 6 < 0 and dividing by 6 yields t 2 5t + < 0 and factoring gives us (t )(t ) < 0 The critical values are and t < t = < t < t = < t t neg 0 pos pos pos t neg neg neg 0 pos (t )(t ) pos 0 neg 0 pos Thus the inequality is true when < t < That is the ball is more than 97 feet above the ground between 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 = ), so we use the sign chart: x < x = < x < 2 x = 2 2 < x < x = < x x + neg 0 pos pos pos pos pos neg neg neg 0 pos pos pos x neg neg neg neg neg 0 pos ()(x + ) x neg 0 pos 0 neg und pos Thus the inequality is true when x 2 or when x > In interval notation, the answer is [, 2] (, ) 6

7 29 The combined variation equation is L = kd To find k we plug in the given values: h2 6 = k2 Therefore, k = 600/6 = 375 Now, if a column has a diameter of 3 feet and a 02 height of feet, it can support up to L = (375)(3 ) tons 30 Increases by a factor of 2, ie, can support a weight 6 times greater 3 Under these parameters, the modified column could support (5) th of the weight of the 32 original column, that is, it can support only of the amount of weight 32 (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 = (05)(25) = 25 Thus the intensity when d = is I = Thus to three decimal places, the sound intensity is 2 003W/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 /9 of the original intensity k (3d) 2 = k 9d = 2 9 k d 2 33 The relation is L = kwd2 l new beam can support We use 600 = k(6)(82 ) L = 700(55)(62 ) 32(6) to find k = ()(600) (6)(6) Thus the new beam can support approximately 27 pounds = 700 Thus the 32 7