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Self-assessment A One-dimensional Mathematics 1. Consider the following numbers: -8 0 0.124373971… 0.23232323…. 5 128 15 4 8 8 15 , 128, 8___ 4 b) Name the irrational numbers: __ 5 , 0.124373971…, a) Name the rational numbers: ___-8, 0, 0.23232323…, c) Name the integers: __-8, 0, 128, 8________________________________ 15 d) Name the real numbers: __ ___-8, 0, 0.23232323…, , 128, 8, 5 , 0.124373971…, 4 e) Name the imaginary numbers: ____ 8 ____________________ f) Name the natural or counting numbers: ___128, 8 ____________________ g) Name the complex numbers: ___All of them are complex______________________ h) Name the whole numbers: ___ 0, 8, 128 _____________________________ 2. Convert the following decimals to an equivalent fraction in reduced form. You can use a calculator. 33 a) 0.033 a) 100 113 ____________ 900 b) 0.12555555…. b) _ c) 0.12424242424242424… c) __ 41 _______________ 330 3. Find the solution of the equation |3 – 5x| = x -3 3. _________________ Solution: |3 – 5x| = x -3 is equivalent to 3-5x = (x-3) or 3-5x = -(x-3) where x – 3 can not be negative. 3 – 5x = x – 3 or 3 – 5x = -x +3 -6x = -6 or -4x = 0 x = 1 or x = 0. But x = 0 does not check because x – 3 = 0 – 3 = -3 Answer: The solution is x = 1 -1- 4. Write each of the following in interval notation; Problem All the real numbers x, such that -3 < x < 7 All the real numbers that are at least -4 but less than 15 All the real numbers between -8 and 30 All the real numbers that are at least 4 All the non-negative real numbers All the real numbers from 5 to 12 Answer [-3, 7) [-4, 15) (-8, 30 ) [ 4, ) [ 0, ) [5, 12] 5. Perform the operations required to simplify and write the answers in the forma a + bi Problem 1. (-3 + 5i) - 3(2 -4i) Solution: -3 + 5i -6 +12i = -9 + 17i 2. (2 + 4i)(-3-5i) Solution: Answer: -9 + 17i 14 – 22i (2 4i)(-3 - 5i) -6 - 10i - 12i - 20i 2 6 22i 20( 1) 6 22i 20 14 22i 3. ( 2 3i ) 3 3 2 ( 2 3i ) ( 2 3i )( 2 3i )( 2 3i ) ( 2 3i )( 4 6i 6i 9i ) -46-9i Solution: ( 2 3i )( 4 12i 9) ( 2 3i )( 5 12i ) 10 24i 15i 36i 2 10 9i 36 46 9i 4. 2i13 5i 34 6i19 6 Solution: 11-4i 2i13 5i 34 6i19 6 2i 5i 2 6i 3 6 2i 5 6i 6 11 4i 5. 20i 7 Solution: 20i 6. 7 20 20 20 20 i 20i 20i 20i 7 3 i i i i2 1 i i 6 8i 2i 0+20i -4 + 3i Solution: 6 8i 2(3 4i ) 3 4i 3 4i i 3i 4i 2 3i 4( 1) 4 3i 2i 2i i i i ( 1) 1 i2 7. 15 10i 2i -8 + i Solution: 15 10i 15 10i 2 i 30 15i 20i 10i 2 40 5i 8 i 2i 2i 2i 5 4 2i 2i i 2 -2- 8. Find the solution of the equation 6x 2 x 2 0 Solution: 6x 2 x 2 0 (3x 2)( 2x 1) 0 3x 2 0 or 2x 1 0 3x 2 or 2x 1 x 2 1 Answer , 3 2 2 1 or x 3 2 9. Solve the equation: Answer: ___x = -1___________- 4 3 7 2 x 2 x 5 x 3x 10 Solution: 4 3 7 4 3 7 , x 2, x 5 x 2 x 5 x 2 3x 10 x 2 x 5 ( x 5)( x 2) 4 3 7 ( x 2)( x 5) ( x 2)( x 5) ( x 2)( x 5) x2 x5 ( x 5)( x 2) 4( x 5) 3( x 2) 7 x 1, it checks 4x 20 3x 6 7 7x 7 Answer : {1} 10. Consider the quadratic equation 4x 2 24x 30 7 a) Give the values of a=4 ___ b=-24 _____ c= 37______ b) Find the value of the discriminant b 2 4ac Answer: -16 b 2 4ac (24) 2 4(4)(37) 576 592 16 c) According to the value of the discriminant, which of the following is true? 1. The equation has two real rational solutions 2. The equation has two equal rational solutions 3. The equation has two irrational solutions 4. The equation has no real solution Answer: 4. The equation has no real solution d) Find the solutions of the equation. b b 2 4ac ( 24) 16 24 4i 24 4 1 x i 3 i 2a 2( 4) 8 8 8 2 Answer : x 3 1 1 i or 3 i 2 2 1 3 11. Find the solution of the inequality 2 1 x . Give answer in interval notation. 3 2 1 3 2 1 x 12 6 2x 9 12 6 2x 9 6 3 2 Solution: 3 3 3 18 2x 3 9 x x 9 Answer : , 9 2 2 2 12. Use the formula x 2 x1 x 2 x x1x 2 0 to find the quadratic equation in each of the following cases. Give answer in the form ax 2 bx c 0 where a, b and c are integers. -3- The roots or solution of the equation are given below 1) -6 and 5 Answer: Equation Solution: x 2 x1 x 2 x x1x 2 0 x 2 x 30 0 Solution: x 2 x1 x 2 x x1x 2 0 x 2 10x 29 0 x 2 6 5x ( 6)(5) 0 x 2 x 30 0 2) 5 – 2i and 5 + 2i x 2 5 2i 5 2i x (5 2i )(5 2i ) 0 x 2 10x 29 0 3) The only solution is -6 Only one solution means that the two solutions are the same: -6, -6 Solution: x 2 x1 x 2 x x1x 2 0 x 2 6 6x (6)( 6) 0 x 2 12x 36 0 2 4 4) The solutions are and 5 3 x 2 12x 36 0 15x 2 14x 8 0 Solution: x 2 x1 x 2 x x1x 2 0 4 2 4 14 8 2 x 2 x 0 x 2 x 0 3 5 3 15 15 5 15x 2 14x 8 0 13. Find the solution of the inequality | 1- 4x | - 7 < -2 . Give answer in interval notation. -2 | 1 -4x | < 5 Solve the equation: 1 – 4x = 5 or 1 – 4x = -5 -4x = 4 or -4x = -6 x = -1 or 6 3 x 1.5 4 2 3 Investigate which interval determined by x = -1 and x 1.5 work 2 1)For x = -2, | 1-4(-2) | -7 = 9 – 7 =2 which is not < -2, therefore the interval , 1 is not a | 1-4x | - 7 < solution. 2)For x = 0, | 1-4(0) | -7 = 1 – 7 =-6 which is < -2, therefore the interval (-1, 1.5) is a solution. 3) 1)For x = 2, | 1-4(2) | -7 = 7 – 7 =0 which is not < -2, therefore the interval 1.5, is not a solution. The solution is the interval [-1, 1.5] -1 1.5 -4-