Practice Problems, Test 3 M102 (1) Simplify: (b) 3 125 81 (a) (c) 4x 2 (d) 3 8x3 (2) Find the distance between the points (-2,4) and (4,-4). (3) Find the missing side of the right triangle: 15 9 b (4) Rewrite the expression without rational exponents, and simplify, if possible: (a) 4 1 2 (b) (8) 1 3 3 2 (c) 25 (5) Simplify the radical expression: (a) 50 (b) 28 (c) 3 40 (d) 3 54 (e) 5 2 18 (6) Multiply the radical expressions: (a) 2( 2 3 3) (b) (2 5 3 2)(3 5 4 2) (c) Multiply 2 3 by it's conjugate. (7) Rationalize the denominator: (a) 3 2 (b) 2 3 3 5 (8) Solve the radical equation: (a) r 25r 144 (b) z 1 z 2 (d) x 4 3 4 (e) 9 5 2 1 1 (f) x 2 x 4 (9) Simplify the square root using complex numbers: (a) 16 (b) 13 (10) Perform the complex operations: (a) (3 4i) (2 5i) (b) 2i (4 3i) (c) (2 5i )(3 6i ) (d) multiply 4 - 3i by it's conjugate. (e) divide: 2 5i 2 3i Solutions: 81 = 9 (since 92 = 81 ) 3 (since 53 = 125) (1) (a) (b) 4 x 2 (2 x) 2 2 x. (c) (d) 125 = 5 3 8 x3 3 (2 x)3 2 x. (2) For this and all such problems, you need to know the distance formula between two points d = ( x1 x0 )2 ( y1 y0 )2 . Here, the points are (-2,4) and (4,-4), so the distance between them is: (4 2) 2 (4 4) 2 62 82 100 10. (3) We use the Pythagorean Theorem to find b a 2 b2 c2 , c a 2 b2 , b c2 a2 . a 2 b2 c2 , c a 2 b2 , b c2 a2 . b 152 92 225 81 144 12. (4) 1 n The crucial rules here are: x n x , m and: x n ( n x )m (remember, for fractional exponents - the numerator is the power, and the denominator is the root). 1 2 (a) 4 4 2. 1 (b) (8) 3 3 8 2. (since (-2)3 = -8) (c) Negative exponents send stuff on the top to the bottom and the bottom to the top: m m 1 1 n x m, xn. m xn x n 25 3 2 1 25 3 2 1 1 1 . 3 ( 25)3 5 125 4 (d) x 3 3 x 4 (The 3 is in the denominator, so it the root, the 4 is in the numerator, and so it is the power - note that with variables, we like to put the power inside of the radical). (e) Here, the negative exponent 'flips' the fraction (since the top goes to bottom and bottom to top - then we take both numbers to the positive version of the exponent: 4 9 5 2 5 2 5 2 9 ( 9)5 35 243 9 5 5 . 5 2 32 ( 4) 4 42 (f) Multiplying same base, different exponents - we have to add the exponents. 1 1 1 1 4 x2 x4 x2 3 x4. (5) Here, we need to pull out some roots. 50 5 2 (a) 25 2 28 2 7 (b) 4 7 3 (c) 3 40 2 3 5 83 5 3 (d) 3 54 3 3 2 27 3 2 (e) Here, we need to simplify one of the radicals (by pulling out roots) , then combine them using the rule: a n x b n x (a b) n x 5 2 18 5 2 3 2 2 2. (6) Radical operations are like polynomial operations - in these two problems, we need to distribute of use FOIL. Note here that I also make use of the fact that a * a a. (a) 2( 2 3 3) 4 3 6 2 3 6. (b) F O I L (2 5 3 2)(3 5 4 2) 6(5) 8 10 6 10 12(2) 6 2 10. (c) The conjugate of 2 3 is 2 3 , and we need to use the difference of squares formula (a + b)(a - b) = a2 - b2. 2 (2 3)(2 3 22 3 4 3 1. (7) Rationalizing the denominator means eliminating the radical in the denominator - the techniques for doing so are different in these two problems - in the first, we just multiply by the square root that appears in the denominator. In the second, we have to multiply by the conjugate of the denominator. (a) 3 3 2 2 2 3 2 2 2 (b) Here is where we have to use the conjugate of the denominator. 2 3 3 5 6 2 5 3 3 15 6 2 5 3 3 15 6 2 5 3 3 15 . 95 4 3 5 3 5 32 ( 5) 2 (8) (a) r 25r 144, ( square) r 2 ( 25r 144) 2 , r 2 25r 144, 25r 144 (quadratic : move everything to one side). r 2 25r 144 0, ( factor :) (r 16)(r 9) 0 ( set factors 0). r 16 0, r 9 0, 36 36 9 9 r 16, r 9. Check the answers: 9 25(9) 144 ? 9 225 144, 9 81. 16 25(16) 144 ? 16 400 144, 16 256. So both candidate solutions work. (b) This is a 'double radical' equation - what you have to do is isolate one radical and square, then simplify, isolate the other radical and square, then solve for the variable. I'm going to isolate z 1 first: z 1 z 2 z 1 2 z z 1 2 ( square) (2 z )(2 z ), z 1 4 2 z 2 z z , ( Isolate z 4 4 z 3 4 z 3 z 4 2 3 4 9 z. 16 Check : z) ( square) z , 2 9 9 1 2? 16 16 25 9 1? 16 16 25 9 1? 16 16 5 3 1. 4 4 Yes. (9) The key idea here is to 'pull the i out' - the square root of -1 is i. (a) 16 16i 4i. (b) 13 13i (10) The key idea with complex operations is to 'treat i as a variable', and thus these operations come down to combining like terms, using distribution and FOIL, and the special property i2 = -1. (a) (3 4i) (2 5i) = 5 + i 3+2 (-4+5)i (combine like terms) (b) 2i (4 3i) = 2i - 4 + 3i = -4 + 5i. (combine like terms) (c) F O I L (2 5i )(3 6i ) = 6 + 12i - 15i - 30i2 -30(-1) 30 = 36 - 3i. (d) The rule for multiplying complex conjugates is slightly different than the regular difference of squares formula, owing to the fact that i2 = -1 alters the simplification: (a bi)(a bi ) a 2 b2 The conjugate of 4 - 3i is 4 + 3i, and (4 3i)(4 3i) 42 32 25. (e) In order to divide by a complex number, we have to multiply by the conjugate of the denominator: 2 5i 2 3i 4 6i 10i 15i 2 4 16i 15(1) 11 16i 11 16 * i. 2 2 2 3i 2 3i 2 3 13 13 13 13

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# Practice Problems, Test 3 M102