6709713-Chapter-01-Quadratic-Equations-in-One-Unknown

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Unformatted text preview: Certificate Mathematics in Action Full Solutions 4A 1 Quadratic Equations in One Unknown Activity Activity 1.1 (p. 28) 1. x x2 x 6 y 4 3 2 1 0 1 2 3 16 9 4 1 0 1 4 9 4 3 2 1 0 1 2 3 6 6 6 6 6 6 6 6 6 0 4 6 6 4 0 6 (2) - (1) , 9 x = 1.5 90 x = 15 15 x = 90 1 = 6 = 1 0.16 6 3. 2. Let x = 0.12, i.e. x = 0.121 212 (1) 100 x = 12.121 212 (2) (2) - (1) , 99 x = 12 12 x = 99 4 = 33 4 0.12 = 33 Let x = 0.123, i.e. x = 0.123 123 1000 x = 123.123 123 4. (1) (2) (2) - (1) , 3. 4. 3, 2 x2 + x - 6 = 0 ( x + 3)( x - 2) = 0 x + 3 = 0 or x - 2 = 0 x = - 3 or 5. x=2 999 x = 123 123 x= 999 41 = 333 41 0.123 = 333 The quadratic equation ax2 + bx + c = 0 can be solved graphically by reading the x-intercepts of the graph of y = ax2 + bx + c. p.6 Quadratic equation (a) (b) (c) (d) (e) (f) p.8 1. (a) x2 10x + 16 = 0 (x 2)(x 8) = 0 x - 2 = 0 or x - 8 = 0 x = 2 or x =8 5x2 = 6 x x2 4 = 5x 3x2 = 4 8x = x2 2 x(3 x) = 0 (1 + x)(1 x) + 3x = 2 General form ax2 + bx + c = 0 5x2 + x 6 = 0 x2 5x 4 = 0 3x2 + 0x 4 = 0 x2 8x + 0 = 0 x2 3x + 2 = 0 x2 3x + 1 = 0 Value of a b c 5 1 6 1 5 4 3 0 4 1 8 0 1 3 2 1 3 1 Follow-up Exercise p. 3 1. Let x = 0.8, i.e. x = 0.888 888 (1) 10 x = 8.888 888 (2) (2) - (1) , 9 x = 8 8 x= 9 = 8 0.8 9 Let x = 0.16, 2. (b) 2x2 + 13x + 15 = 0 (2x + 3)(x + 5) = 0 i.e. x = 0.166 666 (1) 10 x = 1.666 666 (2) 1 1 2x + 3 = 0 or x + 5 = 0 3 x=- or x = -5 2 1. (a) Quadratic Equations in One Unknown x = -1 or x= 1 4 e (c) 3x 2 - 5 x - 2 = 0 (3 x + 1)( x - 2) = 0 3x + 1 = 0 or x - 2 = 0 1 x=- or x=2 3 12 x 2 + x - 6 = 0 ( 4 x + 3)(3x - 2) = 0 4x + 3 = 0 or 3 x - 2 = 0 3 2 x=- or x= 4 3 4 x 2 = 25 4 x 2 - 25 = 0 ( 2 x + 5)(2 x - 5) = 0 2x + 5 = 0 or 2 x - 5 = 0 5 5 x=- or x= 2 2 x ( x + 2) = 15 1 =0 4 x + 1 = 0 or 4 x - 1 = 0 ( x + 1)(4 x - 1) = 0 x + 1 = 0 or x - 4 x 2 + 3x - 1 = 0 2. (a) The required equation is 4 x 2 + 3 x - 1 = 0. 2 1 e x= or x=- (b) 3 2 2 1 x - = 0 or x + = 0 3 2 3x - 2 = 0 or 2 x + 1 = 0 (3x - 2)(2 x + 1) = 0 6x 2 - x - 2 = 0 The required equation is 6x2 x 2 = 0. 2. (a) 2 x 2 - 15 x - 8 = 0 ( 2 x + 1)( x - 8) = 0 2x + 1 = 0 or x - 8 = 0 x=- 1 or 2 x=8 1 1 1 and - 8 2 (b) (b) The roots of the required equation are i.e. - and 2 1 . 8 x= x- 1 8 (c) x 2 + 2 x = 15 x 2 + 2 x - 15 = 0 ( x + 5)( x - 3) = 0 x+5= 0 or x - 3 = 0 x = - 5 or x=3 ( x + 1)( x - 3) = 2( x - 3) x = -2 or x+2=0 or (d) x 2 - 2x - 3 = 2x - 6 x 2 - 4x + 3 = 0 ( x - 1)( x - 3) = 0 x - 1 = 0 or x - 3 = 0 x = 1 or x=3 Alternative Solution ( x + 1)( x - 3) = 2( x - 3) ( x + 1)( x - 3) - 2( x - 3) = 0 ( x - 3)[( x + 1) - 2] = 0 ( x - 3)( x - 1) = 0 x-3= 0 or x -1 = 0 x=3 or x =1 3. (a) 1 =0 8 x + 2 = 0 or 8 x - 1 = 0 ( x + 2)(8 x - 1) = 0 8 x 2 + 15 x - 2 = 0 The required equation is 8x2 + 15x 2 = 0. 4 x 2 - 13 x - 12 = 0 ( 4 x + 3)( x - 4) = 0 4x + 3 = 0 or x - 4 = 0 3 x=- or x=4 4 p.10 3 (b) The roots of the required equation are - 4 and 2 4 3 , i.e. - and 2. 2 8 3 e x = - or x =2 8 3 x + =0 or x -2 = 0 8 8x + 3 = 0 or x 2 = 0 (8 x + 3)( x - 2) = 0 8 x 2 - 13x - 6 = 0 The required equation is 8x2 -13x -6 -0. 2 Certificate Mathematics in Action Full Solutions 4A 1 1 1 x + = x - 2 4 4 1 1 1 x + = x - 3 6 6 2 2 2 2 6. p.13 1. (a) ( x + 1) 2 = 9 x + 1 = 3 x = -1 + 3 or = 2 or ( 2 x - 3) 2 = 2x - 3 = 2x = 3 + 2 5 x = 2 4 2 or or 7. 8. x2 - x2 - x2 + 2 -1 - 3 -4 5 5 5 x + = x + 2 4 4 2 (b) 3-2 1 2 p. 16 x2 - 8x - 9 = 0 x2 - 8x = 9 1. (a) 2. (a) ( x - 2) 2 = 5 x-2 = 5 x = 2+ 2 5 or 2 - 5 8 8 x2 - 8x + = 9 + 2 2 ( x - 4) 2 = 25 x - 4 = 5 x = 45 2 2 (b) x + 1 = 6 4 1 x+ = 6 4 x = 9 or - 1 1 1 x = - + 6 or - - 6 4 4 3. (a) ( x - 3) 2 = 8 x-3= 8 x = 3+ 8 or 3 - 8 (b) 2 2 x - 5x - 3 = 0 5 3 x2 - x - = 0 2 2 5 3 x2 - x = 2 2 2 = 5.83 (cor. to 2. d.p.) or 0.17 (cor. to 2 d.p.) 5 2 (b) ( x + 2) = 4 5 x+2= 4 x+2= 5 2 5 3 5 5 x - x+ = + 2 4 2 4 5 49 x - = 4 16 5 7 x- = 4 4 5 7 x= 4 4 = 3 or - x2 + 7x - 1 = 0 x2 + 7x = 1 7 2 x2 + 7x + = 1 + 2 7 2 2 2 2 2 x = -2 + 5 5 or - 2 - 2 2 = - 0.88 (cor. to 2 d.p.) or - 3.12 (cor. to 2 d.p.) 1 2 p. 15 1. 2. 3. 4. 5. 18 x 2 + 18 x + = ( x + 9) 2 2 12 x 2 - 12 x + = ( x - 6) 2 2 2 2 2. (a) x + 7 = 53 2 4 x+ 7 53 = 2 2 7 53 x = - 2 2 - 7 + 53 - 7 - 53 = or 2 2 2 7 7 x2 + 7x + = x + 2 2 2 2 2 8 x 2 + 8 x + = ( x + 4) 2 2 9 9 x2 - 9x + = x - 2 2 2 2 3 1 3x 2 - 6 x - 2 = 0 2 x2 - 2x - = 0 3 2 x2 - 2x = 3 2 2 2 2 2 x2 - 2x + = + 3 2 2 5 ( x - 1) 2 = 3 5 x -1 = 3 5 x = 1+ or 1 - 3 4. Quadratic Equations in One Unknown Using the quadratic formula, x = = = -b -8 b 2 - 4ac 2a 82 - 4(1)( 3) 2(1) (b) - 8 52 2 - 8 2 13 = 2 = - 4 + 13 or - 4 - 13 5 3 5. 4x = -x 2 + 3 x + 4x - 3 = 0 2 p.20 1. Using the quadratic formula, Using the quadratic formula, x = = = -b -4 x= = = - b b 2 - 4ac 2a - 4 42 - 4(1)(-32) 2(1) b 2 - 4ac 2a 4 2 - 4(1)( 3) 2(1) - 4 144 2 - 4 12 = 2 = 4 or - 8 2. Using the quadratic formula, x = = = -b b 2 - 4ac 2a ( -5) 2 - 4( 2)( 2) 2( 2 ) - 4 28 2 -42 7 = 2 = - 2 + 7 or - 2 - 7 p. 25 1. Quadratic equations Value of x2 + 4x + 2 = 0 = 42 - 4(1)(2) = 8 > 0 2x2 -3x -5 = 0 = (- 2 - 4(2)(- = 49 > 0 3) 5) 2 x + 8x + 16 = 0 = 82 - 4(1)(16) = 0 2x2 + 5x = 0 = 52 - 4(2)(0) = 25 > 0 4x2 + 25 = 0 = 02 - 4(4)(25) = - 400 < 0 Nature of roots 2 distinct 1 double real root No real roots real roots Q The equation x2 -9x + k = 0 has a double real root. = 0 ( -9) 2 - 4(1)( k ) = 0 81 - 4k = 0 81 k = 4 The equation (2k -1)x2 + 3x -6 = 0 has no real roots. - ( -5) 5 9 4 53 = 4 1 = 2 or 2 3x 2 = -10 x + 8 3x 2 + 10 x - 8 = 0 Using the quadratic formula, 2. 3. x= = = - b b - 4ac 2a 2 - 10 10 2 - 4(3)(-8) 2(3) 3. - 10 196 6 - 10 14 = 6 2 = or - 4 3 Q 4 Certificate Mathematics in Action Full Solutions 4A < 0 3 - 4( 2k - 1)( -6) < 0 9 + 48k - 24 < 0 48k - 15 < 0 15 k < 48 5 k < 16 2 (b) 4 x 2 + 36 x = 114 4 x 2 + 36 x - 114 = 0 2 x 2 + 18 x - 57 = 0 Using the quadratic formula, x= = = - b b 2 - 4ac 2a - 18 18 2 - 4( 2)( -57) 2( 2) The range of possible values of k is k < 5 . 16 p. 30 1. The x-intercepts of y = 2x2 + 3x - 5 are - and 1. 2.5 Therefore, the roots of the equation 2x2 + 3x -5 = 0 are - and 1. 2.5 (a) x y - 2 9 - 1 4 0 1 1 0 2 1 3 4 4 9 2. 2. - 18 780 4 - 18 2 195 = 4 - 9 195 = 2 - 9 + 195 - 9 - 195 = or (rejected) 2 2 = 2.48 (cor. to 3 sig. fig.) Let x be the number of days Mr Tung worked in the 3600 project, then Mr Tung's daily wage is \$ and Mr x 3600 - 100 . Pang's daily wage is \$ x 3600 - 100 ( x + 3) = 3600 x (b) 3600 - 100 x + 10 800 - 300 = 3600 x - 100 x 2 + 3300 x + 10 800 = 3600 x 100 x 2 + 300 x - 10800 = 0 (c) The x-intercept of y = x2 -2x + 1 is 1. Therefore, the root of the equation x2 - 2x + 1 = 0 is 1. x 2 + 3x - 108 = 0 ( x + 12)( x - 9) = 0 x + 12 = 0 or x - 9 = 0 x = -12 (rejected) or x = 9 Mr Tung worked in the project for 9 days. p. 33 1. 3. 5. p. 40 1. (a) Area of the border = [( x + 7 + x )( x + 11 + x ) - (7.11)] cm 2 = [( 2 x + 7)( 2 x + 11) - 77] cm = ( 4 x 2 + 36 x + 77 - 77) cm 2 = ( 4 x 2 + 36 x ) cm 2 2 Exercise positive negative negative 2. 4. 6. positive zero positive 1. Exercise 1A (p. 11) Level 1 ( x - 3)( x - 4) = 0 x - 3 = 0 or x - 4 = 0 x = 3 or x = 4 (3x + 2)( x + 2) = 0 3x + 2 = 0 or x+2=0 2 x=- or x = -2 3 3x ( x + 5) = 0 x ( x + 5) = 0 2. 3. 5 1 x=0 x=0 4. Quadratic Equations in One Unknown or or x+5= 0 x = -5 4 y - 3 = 0 or 3 y - 2 = 0 3 2 y = or y = 4 3 12m 2 + 25m + 12 = 0 ( 4m + 3)( 3m + 4) = 0 4m + 3 = 0 or 3m + 4 = 0 3 4 m = - or m = - 4 3 2 - 15 x - 8 x 2 = 0 8 x 2 + 15 x - 2 = 0 (8 x - 1)( x + 2) = 0 8 x - 1 = 0 or x + 2 = 0 1 x = or x = -2 8 5x 2 = 9 x + 2 5x 2 - 9 x - 2 = 0 (5 x + 1)( x - 2) = 0 5x + 1 = 0 or x - 2 = 0 1 x = - or x = 2 5 12 - 4 y = 5 y 2 5 y 2 + 4 y - 12 = 0 (5 y - 6)( y + 2) = 0 5 y - 6 = 0 or y + 2 = 0 6 y = or y = -2 5 x = 0 or x=1 x = 0 or x -1 = 0 x ( x - 1) = 0 x2 - x = 0 The required equation is x2 - x = 0. x = 4 or x=- 1 x -4 = 0 or x + 1 = 0 ( x - 4)( x + 1) = 0 x 2 - 3x - 4 = 0 The required equation is x2 - 3x - 4 = 0. x = 2 or x=3 x -2 = 0 or x -3 = 0 ( x - 2)( x - 3) = 0 x 2 - 5x + 6 = 0 The required equation is x2 - 5x + 6 = 0. x=- 1 or x=- 4 x+1=0 or x + 4 = 0 ( x + 1)( x + 4) = 0 x 2 + 5x + 4 = 0 The required equation is x2 +5x + 4 = 0. 5( x - 1)( x + 3) = 0 ( x - 1)( x + 3) = 0 x - 1 = 0 or x + 3 = 0 x = -3 x = 1 or 2 y2 - 5y = 0 y ( 2 y - 5) = 0 y = 0 or 2 y - 5 = 0 5 y = 0 or y = 2 16 - 121x 2 = 0 121x 2 - 16 = 0 (11x + 4)(11x - 4) = 0 11x + 4 = 0 or 11x - 4 = 0 4 4 x = - or x = 11 11 x2 - 6x + 5 = 0 ( x - 1)( x - 5) = 0 x - 1 = 0 or x - 5 = 0 x = 1 or x = 5 x 2 - 7 x - 30 = 0 ( x + 3)( x - 10) = 0 x+3= 0 or x - 10 = 0 x = - 3 or x = 10 2 x 2 - 15 x - 8 = 0 ( 2 x + 1)( x - 8) = 0 2 x + 1 = 0 or x - 8 = 0 13. 5. 14. 6. 15. 7. 16. 8. 9. 17. Q 18. Q 19. Q 20. Q x=- 10. 1 or 2 x =8 4 x 2 + 13x - 12 = 0 ( 4 x - 3)( x + 4) = 0 4 x - 3 = 0 or x + 4 = 0 3 x = or x = -4 4 6 w 2 + 13w + 6 = 0 (3w + 2)( 2 w + 3) = 0 2w + 2 = 0 or 2 w + 3 = 0 2 3 w = - or w = - 3 2 12 y 2 - 17 y + 6 = 0 ( 4 y - 3)( 3 y - 2) = 0 11. 12. 6 Certificate Mathematics in Action Full Solutions 4A 21. Q x2 + bx - 24 = 0 can be solved using the factor method. x2 + bx - 24 = 0 can be written as ( x - h )( x - k ) = 0 for some real numbers h and k. x 2 - hx - kx + hk = 0 2 y (19 - 4 y ) = 35 26. -8 y 2 + 38 y - 35 = 0 8 y 2 - 38 y + 35 = 0 ( 2 y - 7)( 4 y - 5) = 0 2 y - 7 = 0 or 4 y - 5 = 0 7 5 y = or y = 2 4 15( y 2 + 1) = 34 y 15 y 2 - 34 y + 15 = 0 (3 y - 5)(5 y - 3) = 0 3 y - 5 = 0 or 5 y - 3 = 0 5 3 y= or y= 3 5 ( x - 4)( 6 x + 3) = 3(3 - 2 x ) 28. 6 x 2 - 21x - 12 = 9 - 6 x 6 x 2 - 15 x - 21 = 0 2 x 2 - 5x - 7 = 0 ( 2 x - 7)( x + 1) = 0 2 x - 7 = 0 or x + 1 = 0 7 x = or x = -1 2 5 x ( x - 2 ) = ( x - 2) 2 29. 5 x ( x - 2) - ( x - 2 ) 2 = 0 ( x - 2)[5 x - ( x - 2)] = 0 ( x - 2)( 4 x + 2) = 0 x - 2 = 0 or 4 x + 2 = 0 x = 2 or x = - 1 2 = = = = = y 2 - 36 ( y + 6)( y - 6) 0 0 0 x 2 - ( h + k ) x + hk = 0 By comparing coefficients, we have h + k = b hk = -24 Let h = - k = 6, then b = - + 6 = 2 4, 4 Let h = - k = 8, then b = - + 8 = 5 3, 3 Let h = - k = 12, then b = - + 12 = 10 2, 2 b = 2 or 5 or 10 (or any other reasonable answers) 22. Let h and k be the roots of the equation x2 - 18x + c = 0 27. x = h or x=k x - h = 0 or x - k = 0 ( x - h )( x - k ) = 0 x 2 - ( h + k ) x + hk = 0 By comparing coefficients, we have h + k = 18 hk = c Let h = 6, k = 12, then c = 6 12 = 72 Let h = 7, k = 11, then c = 7 11 = 77 Let h = 8, k = 10, then c = 8 10 = 80 c = 72 or 77 or 80 (or any other reasonable answers) Level 2 23. ( x + 3)( 2 x + 1) + 3 = 0 2x2 + 7x + 6 = 0 ( 2 x + 3)( x + 2) = 0 2x + 3 = 0 or x + 2 = 0 3 x = - or x = -2 2 (3x + 1)(1 - 2 x ) + 4 = 0 24. - 6x 2 + x + 5 = 0 6x 2 - x - 5 = 0 ( 6 x + 5)( x - 1) = 0 6x + 5 = 0 or x - 1 = 0 5 x = - or x =1 6 x 2 - 3x ( x - 5) = 25 25. - 2 x 2 + 15 x - 25 = 0 2 x 2 - 15 x + 25 = 0 ( 2 x - 5)( x - 5) = 0 2 x - 5 = 0 or x - 5 = 0 5 x = or x = 5 2 30. 3 y ( y - 6) 3 y ( y - 6) 3 y ( y - 6) - ( y + 6)( y - 6) ( y - 6)[3 y - ( y + 6)] ( y - 6)( 2 y - 6) y - 6 = 0 or 2 y - 6 = 0 y = 6 or y = 3 ( 2 x - 5) 2 = ( x - 3)( 2 x - 5) 31. ( 2 x - 5) 2 - ( x - 3)( 2 x - 5) = 0 ( 2 x - 5)[( 2 x - 5) - ( x - 3)] = 0 ( 2 x - 5)( x - 2) = 0 2 x - 5 = 0 or x - 2 = 0 5 x = or x = 2 2 ( x + 1) 2 + 3( x + 1) - 4 = 0 [( x + 1) - 1][( x + 1) + 4] = 0 x ( x + 5) = 0 32. 7 1 x = 0 or x + 5 = 0 x = 0 or x = -5 x = 33. Q x - 1 or 2 x = -2 38. (a) Quadratic Equations in One Unknown ( 2 x - 3)( 2 x - 1) = 0 4 x 2 - 8x + 3 = 0 The required equation is 4x2 -8x + 3 = 0. 1 = 0 or x + 2 = 0 2 2 x - 1 = 0 or x + 2 = 0 ( 2 x - 1)( x + 2) = 0 2 x 2 + 3x - 2 = 0 The required equation is 2x2 + 3x -2 = 0. x = - 1 or 3 x = 3 2 4x 2 - 8x + 3 = 0 ( 2 x - 1)( 2 x - 3) = 0 2 x - 1 = 0 or 2 x - 3 = 0 1 3 x = or x = 2 2 34. Q 1 (b) The roots of the required equation are 2 and 2 3 2 , i.e. 1 and 3. 2 1 3 = 0 or x - = 0 3 2 3x + 1 = 0 or 2 x - 3 = 0 (3x + 1)( 2 x - 3) = 0 x + 6x2 - 7x - 3 = 0 The required equation is 6x2 - 7x - 3 = 0. 1 1 x = - or x = - 2 3 1 1 x + = 0 or x + = 0 2 3 2x + 1 = 0 or 3x + 1 = 0 ( 2 x + 1)(3 x + 1) = 0 6 x 2 + 5x + 1 = 0 The required equation is 6x2 + 5x + 1 = 0. 1 2 7 3 x= or x=- 3 2 7 3 x - = 0 or x+ =0 3 2 3 x - 7 = 0 or 2 x + 3 = 0 (3x - 7)( 2 x + 3) = 0 x=2 x = -1 6 x 2 - 5 x - 21 = 0 The required equation is 6x2 -5x -21 = 0. 3x 2 - 8 x + 4 = 0 (3x - 2)( x - 2) = 0 3x - 2 = 0 or x - 2 = 0 2 x = or x = 2 3 1 or 3 39. (a) x =1 or x=3 x - 1 = 0 or x - 3 = 0 ( x - 1)( x - 3) = 0 x2 - 4x + 3 = 0 The required equation is x2 -4x + 3 = 0. 2x2 - 9x - 5 = 0 ( 2 x + 1)( x - 5) = 0 2x + 1 = 0 or x - 5 = 0 1 x = - or x = 5 2 1 + 5 and 2 35. Q (b) The roots of the required equation are - - 36. Q 37. (a) 1 9 5 (5) , i.e. and - . 2 2 2 9 5 x= or x =- 2 2 9 5 x - =0 or x + =0 2 2 2x - 9 = 0 or 2x + 5 = 0 ( 2 x - 9)( 2 x + 5) = 0 4 x 2 - 8 x - 45 = 0 The required equation is 4x2 -8x -45 = 0. Exercise 1B (p. 17) Level 1 ( x + 5) 2 = 36 x + 5 = 6 x = -5 + 6 or - 5 - 6 =1 or - 11 1. (b) The roots of the required equation are 3 1 and . 2 2 1 2 3 and 1 , i.e. 2 x= 3 or 2 x= 1 2 3 1 = 0 or x - = 0 2 2 2 x - 3 = 0 or 2 x - 1 = 0 x- 8 Certificate Mathematics in Action Full Solutions 4A ( x - 4) 2 = 2. 16 9 4 x-4= 3 x =4+ 4 4 or 4 - 3 3 16 8 = or 3 3 7. 2 9 2 x - = 5 3 2 2x - 2 = 5 3 9 2 2x - 3. x + 1 = 25 3 1 x + = 5 3 1 1 x = - + 5 or - - 5 3 3 14 16 = or - 3 3 2 2 5 = 3 3 2 5 2x = 3 2+ 5 2- 5 x = or 6 6 2 x 25 - 2 = 7 3 7 x - 2 = 25 3 8. 2 4. 1 25 x - = 4 5 1 4 x- = 5 25 1 2 x- = 5 5 1 2 1 2 x = + or - 5 5 5 5 3 1 = or - 5 5 2 2 x 7 -2 = 3 5 x 7 = 2 3 5 3 7 3 7 x = 6+ or 6 - 5 5 6 p= 2 =9 2 9. 6 x2 - 6x + 9 = x - 2 = ( x - 3) 2 2 2 ( x - 3) 2 = 2 5. x-3= 2 x = 3+ 2 or 3 - 2 8 p = 10. 2 = 16 6. 4( x + 5) 2 = 7 7 ( x + 5) 2 = 4 x+5= 7 2 7 7 or - 5 - 2 2 8 x 2 + 8 x + 16 = x + 2 = ( x + 4) 2 2 2 x = -5+ 5 p = 2 11. 25 = 4 x 2 + 5x + 25 5 = x + 4 2 2 3 p = 4 12. 9 = 16 2 x2 - 3x 9 3 + = x - 2 16 4 2 9 1 Quadratic Equations in One Unknown 13. Q x2 - 2ax + b is a perfect square. 2 - 2a b = 2 = a2 a = 2, b = 4 or a = 1, b = 1 or a = - b = 1. 1, (or any other reasonable answers) ( x - d )2 = 14. 1 4 1 2 1 2 18. 2 x 2 - 10 x - 5 = 0 5 x 2 - 5x - = 0 2 5 x 2 - 5x = 2 2 2 5 5 5 x 2 - 5x + = + 2 2 2 x - 5 = 35 2 4 x- 5 35 = 2 2 5 35 x = 2 2 5 + 35 5 - 35 = or 2 2 2 x-d = x = d e The roots of (x - d)2 = 1 are integers. 4 1 1 1 d = 1 or 2 or 3 (or any other reasonable 2 2 2 answers) Level 2 x 2 - 8 x + 15 = 0 x 2 - 8 x = -15 15. 8 x 2 - 8x + 2 ( x - 4) 2 x-4 x 2 19. 2 8 = -15 + 2 =1 = 1 = 4 1 = 5 or 3 3x 2 + 6 x - 28 = 0 28 x2 + 2x - = 0 3 28 x2 + 2x = 3 2 2 2 28 2 x2 + 2x + = + 3 2 2 31 ( x + 1) 2 = 3 31 x +1= 3 31 = -1+ or - 1 - 3 3x 2 - 6 x - 7 = 0 7 x2 - 2x - = 0 3 7 x2 - 2x = 3 2 2 2 2 = 7 + 2 x - 2x + 3 2 2 10 2 ( x - 1) = 3 10 x -1 = 3 10 x = 1+ or 1 - 3 31 3 x - 16 x + 28 = 0 2 x 2 - 16 x = -28 16. 16 x 2 - 16 x + 2 ( x - 8) 2 x-8 x 2 16 = -28 + 2 = 36 = 6 = 86 = 14 or 2 2 20. x 2 + 7 x - 98 = 0 x 2 + 7 x = 98 7 7 x 2 + 7 x + = 98 + 2 2 2 2 2 10 3 21. 3x 2 + 7 x - 3 = 0 x2 + 7 x -1 = 0 3 7 x2 + x = 1 3 2 2 17. 7 441 x + = 2 4 7 21 x+ = 2 2 7 21 x=- 2 2 = 7 or - 14 x2 + 7 7 7 x + = 1+ 3 6 6 7 85 x+ = 6 36 2 10 Certificate Mathematics in Action Full Solutions 4A 7 85 = 6 6 7 85 x=- 6 6 - 7 + 85 - 7 - 85 = or 6 6 x+ x 2 + 2 3x + 1 = 0 x 2 + 2 3 x = -1 2 3 2 3 x 2 + 2 3x + 2 = -1 + 2 ( x + 3)2 = 2 x+ 3 = 2 x=- 3 2 = - 3 + 2 or - 3 - 2 3 = 0 4 3 x 2 - 2 5x - = 0 2 3 x 2 - 2 5x = 2 5x - 25. 3 2 5 2 5 x 2 - 2 5x + + = 2 2 2 13 (x - 5)2 = 2 13 x - 5 = 2 13 = 5 + or 2 2 2 2 2 24. 22. 5( x - 1) = 4 x 5x2 - 4x - 5 = 0 2 4 x2 - x - 1 = 0 5 4 x2 - x = 1 5 x2 - 4 4 4 x + = 1+ 5 10 10 2 29 x - = 5 25 x- 2 29 = 5 5 2 29 x= 5 5 2 + 29 2 - 29 = or 5 5 2 2 2 x2 - 2 5 - 13 2 2 x 2 - 5 x - 30 = 0 x2 - 23. ( x -1)(2 +3 x ) = x 2 3x 2 - x - 2 = x 2 2x 2 - x - 2 = 0 1 x2 - x -1= 0 2 1 x2 - x =1 2 x2 - 1 1 1 x + =1+ 2 4 4 1 17 x - = 4 16 1 17 x- = 4 4 1 17 x= 4 4 1 + 17 1 - 17 = or 4 4 2 2 2 5 x - 15 = 0 2 5 x2 - x = 15 2 2 2 x2 - 26. 5 5 5 x+ 4 = 15 + 4 2 x - 5 = 245 4 16 x- 5 7 5 = 4 4 5 7 5 x= 4 4 = 2 5 or - 2 3 5 2 11 1 Exercise 1C (p. 21) Level 1 1. Using the quadratic formula, x = = = -b b 2 - 4ac 2a ( -4) 2 - 4(1)( -5) 2(1) 5. Quadratic Equations in One Unknown Using the quadratic formula, x= = = - b b 2 - 4ac 2a - ( -7) (-7) 2 - 4(1)(11) 2(1) - ( -4 ) 4 36 2 46 = 2 = 5 or - 1 2. Using the quadratic formula, 7 5 2 7+ 5 7- 5 = or 2 2 6. Using the quadratic formula, x= = = - b b 2 - 4ac 2a - 5 5 2 - 4(2)(-6) 2(2) x= = = - b b 2 - 4ac 2a - (-3) (-3) 2 - 4(2)(-2) 2( 2) 7. - 5 73 4 - 5 + 73 - 5 - 73 = or 4 4 Using the quadratic formula, 3 25 4 35 = 4 1 = 2 or - 2 3. Using the quadratic formula, x= = = - b b 2 - 4ac 2a - (-2) (-2) 2 - 4(3)(-13) 2(3) x= = = - b b 2 - 4ac 2a - (-9) (-9) 2 - 4(35)(-2) 2(35) 8. 9 361 70 9 19 = 70 2 1 = or - 5 7 4. Using the quadratic formula, 2 160 6 2 4 10 = 6 1 + 2 10 1 - 2 10 = or 3 3 Using the quadratic formula, x= = = - b b 2 - 4ac 2a - 3 3 2 - 4(3)( -5) 2(3) x= = = - b b 2 - 4ac 2a - ( -31) (-31) 2 - 4(14)(15) 2(14) - 3 69 6 - 3 + 69 - 3 - 69 = or 6 6 31 121 28 31 11 = 28 3 5 = or 2 7 12 Certificate Mathematics in Action Full Solutions 4A 9. Using the quadratic formula, 13. (3x + 2)(3x - 2) + 3x = 0 9 x 2 + 3x - 4 = 0 Using the quadratic formula, x = = = -b -3 b 2 - 4ac 2a 32 - 4(9)( -4) 2(9) - b b 2 - 4ac x= 2a = - (-6) (-6) 2 - 4(2)(3) 2( 2) 6 12 = 4 62 3 = 4 3+ 3 3- 3 = or 2 2 10. Using the quadratic formula, - 3 153 18 - 3 3 17 = 18 - 1 + 17 - 1 - 17 = or 6 6 = 0.52 (cor. to 2 d.p.) or - 0.85 (cor. to 2 d.p.) x ( 3x + 4) = ( x + 1)( x - 2) 3x 2 + 4 x = x 2 - x - 2 2 x 2 + 5x + 2 = 0 Using the quadratic formula, x= = = - b b 2 - 4ac 2a - 5 5 2 - 4( 2)(2) 2( 2) x= = = - b b 2 - 4ac 2a - (-7) (-7)2 - 4(6)(-1) 2(6) 14. 7 73 12 7 + 73 7 - 73 = or 12 12 Level 2 11. x ( x - 10) + 5 = 0 x 2 - 10 x + 5 = 0 Using the quadratic formula, x = = = -b b 2 - 4ac 2a ( -10) 2 - 4(1)(5) 2(1) -5 9 4 -53 = 4 = - 0.5 or - 2 - ( -10) 10 80 2 10 4 5 = 2 = 5+2 5 or 5 - 2 5 = 9.47 (cor. to 2 d.p.) or 0.53 (cor. to 2 d.p.) x + 1 ( x + 1) = - 3 x 2 4 3 1 3x 2 x + x+ = - 15. 2 2 4 9x 1 2 x + + = 0 4 2 2 4x + 9x + 2 = 0 Using the quadratic formula, x= = = - b b 2 - 4ac 2a - 9 92 - 4(4)(2) 2( 4) 12. 2 5x + 2 = 7 x 2 7 x - 5x - 2 = 0 Using the quadratic formula, x = = = -b b 2 - 4ac 2a ( -5) 2 - 4(7)( -2) 2( 7) - 9 49 8 -97 = 8 = - 0.25 or - 2 ( x - 3)( 2 x - 5) = x ( x + 3) + 2 2 x 2 - 11x + 15 = x + 3 x + 2 x 2 - 14 x + 13 = 0 Using the quadratic formula, 2 - ( -5) 5 81 14 59 = 14 2 7 = 1 or - 0.29 (cor. to 2 d.p.) = 1 or - 16. 13 1 -b b 2 - 4ac 2a ( -14) - 4(1)(13) 2(1) 2 Quadratic Equations in One Unknown x = = = 20. ( 2 x + 1)( 3x - 2) = (6 x + 5) 2 - 4 x 6 x 2 - x - 2 = 36 x 2 + 56 x + 25 30 x 2 + 57 x + 27 = 0 Using the quadratic formula, x = = = -b - 57 b 2 - 4ac 2a 57 2 - 4(30)( 27) 2(30) - ( -14) 14 144 2 14 12 = 2 = 13 or 1 ( 2 x - 3) 2 + 6 x = 7 4x2 - 6x + 9 = 7 4x2 - 6x + 2 = 0 Using the quadratic formula, x = = = -b b 2 - 4ac 2a ( -6) 2 - 4( 4)( 2) 2( 4) 17. - 57 9 60 - 57 3 = 60 = - 0.9 or - 1 Exercise 1D (p. 26) - ( -6) 6 4 8 62 = 8 = 1 or 0.5 0.2 x 2 + 0.8( x + 2) = 0.8 18. 0.2 x 2 + 0.8 x + 1.6 = 0.8 0.2 x + 0.8 x + 0.8 = 0 2 Level 1 1. For x 2 - 4 x - 1 = 0, = ( -4) 2 - 4(1)( -1) = 20 >0 The equation x2 - 4x - 1= 0 has two distinct real roots. x2 + 4x + 4 = 0 Using the quadratic formula, x = = = -b -4 b 2 - 4ac 2a 4 - 4(1)( 4) 2(1) 2 2. For 2 x 2 + 3x + 1 = 0, = 32 - 4( 2)(1) =1 >0 The equation 2x2 + 3x + 1 = 0 has two distinct real roots. -4 2 = -2 2 0 19. ( x - 3) = ( 2 x + 1)( x - 2) x - 6 x + 9 = 2 x 2 - 3x - 2 2 x 2 + 3x - 11 = 0 Using the quadratic formula, x= - b b 2 - 4ac 2a 3. For 3x2 + 2x + 2 = 0 = 2 2 - 4(3)( 2) = - 20 <0 The equation 3x2 + 2x + 2 = 0 has no real roots. - 3 32 - 4(1)(-11) = 2(1) - 3 53 2 - 3 53 - 3 - 53 = or 2 2 = 2.14 (cor. to 2 d.p.) or - 5.14 (cor. to 2 d.p.) = 4. For x2 + 4x + 4 = 0, 14 Certificate Mathematics in Action Full Solutions 4A = 4 2 - 4(1)( 4) = 0 =0 The equation x2 + 4x + 4 = 0 has a double real root. roots. >0 (-3) - 4(2)(k ) > 0 2 9 - 8k > 0 9 k< 8 The range of possible values of k is k < 9 . 8 5. For 2x2 - 4x + 3 = 0, = ( -4) 2 - 4( 2)(3) = -8 <0 The equation 2x2 - 4x + 3 = 0 has no real roots. 10. Q The equation 5x2 + 3x + (k + 1) = 0 has two distinct real roots. >0 3 - 4(5)(k + 1) > 0 9 - 20k - 20 > 0 - 11 - 20k > 0 2 6. For 3x + x - 10 = 0, 2 k<- 11 20 11 . 20 = 12 - 4(3)(-10) = 121 >0 The equation 3x2 + x - 10 = 0 has two distinct real roots. The range of possible values of k is k < - 11. Q The quadratic equation 3x2 + 4x + k = 0 has real roots. 7. Q The equation (k - 1)x2 - 2x + 1 = 0 has a double real root. = 0 ( -2) - 4( k - 1)(1) 4 - 4k + 4 8 - 4k k 2 0 16 - 12k 0 i.e. 42 - 4(3)(-k ) 0 = = = = 0 0 0 2 12k 16 4 k 3 The range of the values of k is k 4 . 3 8. Q The equation x2 + 4kx + 16k + 20 = 0 has a double real root. = 0 ( 4k ) 2 - 4(1)(16k + 20) = 0 16k 2 - 64k - 80 = 0 k 2 - 4k - 5 = 0 ( k - 5)( k + 1) = 0 k - 5 = 0 or k + 1 = 0 k = 5 or k = -1 12. Q The quadratic equation (k + 1) x2 - 2kx + (k - 2) = 0 has real roots. i.e. 0 (-2k ) 2 - 4( k + 1)(k - 2) 0 4 k 2 - 4k 2 + 4k + 8 0 4k + 8 0 4 k -8 k -2 The range of the values of k is k - 2. 9. Q The equation 2x2 - 3x + k = 0 has two distinct real 15 1 13. Q The equation 3x2 + 5x - k = 0 has no real roots. < 0 5 - 4(3)( - k ) < 0 25 + 12k < 0 2 Quadratic Equations in One Unknown e The equation (k + 2) x2 - 2x + 1 = 0 has no real roots. < 0 25 k < - 12 The range of possible values of k is k < - 25 . 12 18. 14. Q The equation (k + 2)x2 - 4x - 5 = 0 has no real roots. < 0 ( -4) 2 - 4( k + 2)( -5) < 0 16 + 20k + 40 < 0 56 + 20k < 0 k < - 14 5 14 . 5 ( -2) - 4( k + 2)(1) < 0 4 - 4k - 8 < 0 k > -1 2 The range of possible values of k is k > - 1. 3( x 2 + 1) = x + 12k 3 x 2 + 3 = x + 12k 3 x 2 - x + (3 - 12k ) = 0 e The equation 3x2 -x + (3 -12k) = 0 has no real roots. <0 ( -1) - 4(3)(3 - 12k ) < 0 1 - 36 + 144k < 0 35 k< 144 2 The range of possible values of k is k < - 15. Q The equation ax + 3x + c = 0 has two distinct real roots. > 0 2 The range of possible values of k is k < 35 . 144 19. (a) Q The equation x2 - 2kx + 2k + 15 = 0 has equal real roots. 32 - 4ac > 0 9 - 4ac > 0 9 ac < 4 a= 1 , c = 3 or a = 1, c = 2. 2 =0 (-2k ) - 4(1)(2k + 15) = 0 2 4k 2 - 8k - 60 = 0 k 2 - 2k - 15 = 0 (k + 3)(k - 5) = 0 k +3= 0 or k - 5 = 0 k = - 3 or k = 5 (or any other reasonable answers) 16. ax 2 + 2 x = c ax 2 + 2 x - c = 0 e The equation ax2 + 2x - c = 0 has no real roots. < 0 2 2 - 4a ( - c ) < 0 4 + 4ac < 0 ac < -1 ac = - or - (or any other reasonable answers) 2 3. (b) For k = - 3, x 2 - 2( -3) x + 2( -3) + 15 = 0 x 2 + 6x + 9 = 0 ( x + 3) 2 = 0 x +3= 0 x = -3 For k = 5, Level 2 17. ( k + 2) x 2 = 2 x - 1 ( k + 2) x 2 - 2 x + 1 = 0 16 Certificate Mathematics in Action Full Solutions 4A x 2 - 2(5) x + 2(5) + 15 = 0 x 2` - 10 x + 25 = 0 ( x - 5) 2 = 0 x -5 = 0 x = 5 (m - 1) 2 0 - 4 (m - 1) 2 + 1 < 0 <0 [ (m - 1) 2 + 1 > 0 20. (a) Q The equation 4x2 + 5kx + k = 0 has equal real roots. = 0 (5k ) 2 - 4( 4)( k ) = 0 25k 2 - 16k = 0 k ( 25k - 16) = 0 22. (a) The equation x2 -2(m + 1)x + (2m2 + 3) = 0 has no real roots for any real values of m. 12 x 2 + 4 x + k = 1 12 x + 4 x + ( k - 1) = 0 2 e The equation 12x2 + 4x + (k - 1) = 0 has real roots. 0 k = 0 or 25k - 16 = 0 k = 0 or (b) For k = 0, 4 x 2 + 5(0) x + (0) = 0 4x2 = 0 x = 0 16 , 25 16 k= 25 4 2 - 4(12)( k - 1) 0 16 - 48k + 48 0 64 - 48k 0 4 k 3 The range of possible values of k is k 4 . 3 For k = (b) (i) 12 x 2 + 4 x - 1 = 0 12 x 2 + 4 x + 0 = 1 e k = 0 4 3 The equation 12x2 + 4x -1 = 0 has real roots. 16 16 4 x 2 + 5 x + = 0 25 25 16 16 4x2 + x+ = 0 5 25 2 25 x + 20 x + 4 = 0 (5 x + 2 ) 2 = 0 5x + 2 = 0 x = - 2 5 (ii) 21. For the equation x 2 - 2( m + 1) x + ( 2m 2 + 3) = 0 , = [ -2( m + 1)]2 - 4(1)( 2m 2 + 3) = 4m 2 + 8m + 4 - 8m 2 - 12 = -4m 2 + 8m - 8 = -4 ( m 2 - 2 m + 2 ) = -4[( m - 1) + 1] 2 36 x 2 + 12 x + 1 = 0 1 12 x 2 + 4 x + = 0 3 4 12 x 2 + 4 x + =1 3 k = 4 3 4 3 e The equation 36x2 + 12x + 1 = 0 has real roots. For any real values of m, Exercise 1E (p. 34) 17 1 Quadratic Equations in One Unknown Level 1 1. (a) two (b) The x-intercepts of y = x2 -14x + 48 are 6 and 8. Therefore, the roots of the equation x2 - 14x + 48 = 0 are 6 and 8. 2. (a) two (b) The x-intercepts of y = x2 -3x -10 are - and 5. 2 Therefore, the roots of the equation x2 - 3x - = 0 10 are - and 5. 2 (c) The x-intercepts of y = 2x2 - 4x are 0 and 2.0. Therefore, the roots of the equation 2x2 - = 0 are 4x 0 and 2.0. 3. (a) x y - 4 5 - 3 0 - 2 - 3 - 1 - 4 0 - 3 1 0 2 5 x y - 3 14 - 2 5 - 1 0 0 - 1 1 2 2 9 3 20 5. (a) (b) (b) The x-intercepts of y = 2x2 + x - are - and 0.5. 1 1.0 Therefore, the roots of the equation 2x2 + x -1 = 0 are - and 0.5. 1.0 6. (c) The x-intercepts of y = x + 2x - are - and 1. 3 3 2 (a) x y 0 9 1 4 2 1 3 0 4 1 5 4 6 9 Therefore, the roots of the equation x2 + 2x - 3 = 0 are - and 1.0. 3.0 x 0 - 2 - 1 y 16 6 0 1 - 2 2 0 3 6 4 16 4. (a) (b) 18 Certificate Mathematics in Action Full Solutions 4A Level 2 9. (a) x y - 3 8 - 2 2 - 1 - 2 0 - 4 1 - 4 2 - 2 3 2 4 8 (b) The x-intercept of y = x2 -6x + 9 is 3.0. Therefore, the root of the equation x2 - 6x + 9 = 0 is 3.0. 7. Q The graph y = - 2 + bx + c does not intersect the x x-axis. (b) x2 = x + 4 x - x-4 = 0 2 The equation - 2 + bx + c = 0 has no real roots. x < 0 b 2 - 4( -1)( c ) < 0 b 2 + 4c < 0 The x-intercepts of y = x2 - x - 4 read from the graph are - and 2.6. 1.6 Therefore, the roots of the equation x2 = x + 4 are 1.6 - and 2.6. x 10. (a) y - 3 - 5 - 2 - 2 - 1 - 1 0 - 2 1 2 3 - - - 5 10 17 The possible values of b and c are: b = 1, c = - or 2 b = 0, c = - (or any other reasonable answers) 1. 8. Q The graph y = kx2 + 3x + k cuts the x-axis at two distinct points. The equation kx2 + 3x + k = 0 has 2 distinct real root. >0 3 - 4( k )(k ) > 0 2 9 - 4k 2 > 0 9 k2 > 4 3 3 - <k< 2 2 k = - or 1 1 or 1 (or any other reasonable answers) 2 (b) The graph y = - 2 -2x -2 does not intersect the x x-axis. Therefore, the equation - 2 - - = 0 has no real x 2x 2 roots. 19 1 (b) (i) 11. (a) x y - 1 9 0 1 1 1 2 9 Quadratic Equations in One Unknown x y 0 - 5 1 2 3 4 - 1.8 - 0.2 - 0.2 - 1.8 5 - 5 (ii) The x-intercept of y = (k + 1)x2 + 4x -5 is 2.5. Therefore, the root of the equation (b) 4 x (1 - x ) = 1 4x2 - 4x + 1 = 0 The x-intercept of y = 4x2 -4x + 1 is 0.5. Therefore, the root of the equation 4x(1 - x) = 1 is 0.5. (k + 1)x2 + 4x - 5 = 0 is 2.5. Exercise 1F (p. 40) Level 1 12. The graph y = x2 -2x -k does not intersect the x-axis. Therefore, the equation x2 - 2x - k = 0 has no real roots. < 0 ( -2) - 4(1)( -k ) < 0 4 + 4k < 0 k < -1 2 1. Let x be the positive number. x + x 2 = 56 x + x - 56 = 0 ( x + 8)( x - 7) = 0 2 The range of possible values of k is k < - 1. x+8 = 0 or x - 7 = 0 x = -8 (rejected) or x = 7 The positive number is 7. 13. (a) The graph of y = (k + 1)x2 + 4x - touches the 5 x-axis. Therefore, the equation (k + 1)x2 + 4x -5 = 0 has a double real roots. 2. Let x be the smaller odd number, then x + 2 is the larger odd number. =0 x 2 + ( x + 2) 2 = 202 x 2 + x 2 + 4 x + 4 = 202 2 x 2 + 4 x - 198 = 0 x 2 + 2 x - 99 = 0 ( x + 11)( x - 9) = 0 x + 11 = 0 or x - 9 = 0 x = -11 (rejected) or x = 9 The two consecutive positive odd numbers are 9 42 - 4( k + 1)(-5) = 0 16 + 20( k + 1) = 0 20k + 36 = 0 k=- 9 5 20 Certificate Mathematics in Action Full Solutions 4A and 11. 150 2 x + = 50 x 150 x+ = 25 x x 2 + 150 = 25 x x 2 - 25 x + 150 = 0 ( x - 15)(( x - 10) = 0 x - 15 = 0 or x = 15 or x - 10 = 0 x = 10 3. Let x be the smallest number. [ x + ( x + 1) + ( x + 2) 2 ] - [ x 2 + ( x + 1) 2 + ( x + 2) 2 ] = 94 (3 x + 3) 2 - [ x 2 + x 2 + 2 x + 1 + x 2 + 4 x + 4] = 94 9 x 2 + 18 x + 9 - (3 x 2 + 6 x + 5) = 94 6 x 2 + 12 x + 4 = 94 6 x 2 + 12 x - 90 = 0 x 2 + 2 x - 15 = 0 ( x + 5)( x - 3) = 0 The length and width are 15 cm and 10 cm respectively. x+5=0 or x - 3 = 0 x = -5 (rejected) or x = 3 The three positive consecutive numbers are 3, 4 and 5. 4. Let x cm be the length of the hypotenuse, then (x -2) cm is the base and [(x - 2) - 14] = (x - 16) cm is the height. ( x - 2) 2 + ( x - 16) 2 = x 2 (Pyth. theorem) x 2 - 4 x + 4 + x 2 - 32 x + 256 = x 2 x 2 - 36 x + 260 = 0 ( x - 26)( x - 10) = 0 x - 26 = 0 or x = 26 or x - 10 = 0 x = 10 (rejected ) Alternative Solution 50 - 2 x Let x cm be the length, then = ( 25 - x ) cm 2 is the width. x ( 25 - x ) = 150 2 25 x - x 2 = 150 x - 25 x + 150 = 0 ( x - 15)( x - 10) = 0 x - 15 = 0 or x = 15 or x - 10 = 0 x = 10 The length and width are 15 cm and 10 cm respectively. The length of the hypotenuse is 26 cm. 5. Let x m be the width, then (x + 2) m is the length. x + ( x + 2) = 10 (Pyth. theorem) 2 2 2 7. Let x cm be the width of the border. Area of the border = 1032 cm2 ( 42 + 2 x )( 32 + 2 x ) - 42 32 = 1032 4 x 2 + 148 x + 1344 - 1344 - 1032 = 0 4 x 2 + 148 x - 1032 = 0 x 2 + 37 x - 258 = 0 ( x + 43)( x - 6) = 0 x + x + 4 x + 4 = 100 2 2 2 x 2 + 4 x - 96 = 0 x 2 + 2 x - 48 = 0 ( x + 8)( x - 6) = 0 x+8 = 0 or x = -8 (rejected) or x-6 = 0 x = 6 x + 43 = 0 The length is 8 m and the width is 6 m. or x - 6 = 0 x = -43 (rejected) or x = 6 The width of the border is 6 cm. 6. Let x cm be the length, then 150 cm is the width. x 8. 1 + 9x - 2x2 = 5 2x2 - 9x + 4 = 0 ( 2 x - 1)( x - 4) = 0 x-4 = 0 x = 4 2 x - 1 = 0 or x = 0.5 or 21 1 After 0.5 seconds and 4 seconds, the ball is 5 m above the ground. Quadratic Equations in One Unknown or x + 20 = 0 x - 100 = 0 x = -20 (rejected) x = 100 or The man bought 100 articles. Level 2 12. (a) 2 Base area = ( 26 - 2 x )(16 - 2 x ) cm = ( 4 x 2 - 84 x + 416) cm 2 9. Let x km/h be the original speed of Peter. 4 x 2 - 84 x + 416 = 200 (b) 4 x 2 - 84 x + 216 = 0 x 2 - 21x + 54 = 0 ( x - 18)( x - 3) = 0 x - 18 = 0 or x - 3 = 0 x = 18 (rejected) or x = 3 Volume of the box = base area height of the box = 200 x cm 3 = 200(3) cm 3 = 600 cm 3 20 20 - =1 x -1 x 20 x - 20( x - 1) =1 x( x - 1) 20 x - 20 x + 20 = x 2 - x x - x - 20 = 0 ( x - 5)( x + 4) = 0 2 x - 5 = 0 or x = 5 or x+4 = 0 x = -4 (rejected) The original speed of Peter is 5 km/h. 10. Let \$x be the original price of an apple. 75 75 - =5 x - 0.5 x 75 x - 75( x - 0.5) =5 x( x - 0.5) 75 x - 75 x + 37.5 = 5 x 2 - 2.5 x 5 x 2 - 2.5 x - 37.5 = 0 2 x - x - 15 = 0 ( 2 x + 5)( x - 3) = 0 2 13. (a) (i) Distance travelled in the first x hours = x ( x + 1) km = ( x 2 + x ) km (ii) Distance travelled in the last (x - 1) hours = ( x - 1)( x - 1) km = ( x 2 - 2 x + 1) km 2x + 5 = 0 or x -3 = 0 x=3 ( x 2 + x) + ( x 2 - 2 x + 1) = 16 (b) 5 x = - (rejected) or 2 2 x 2 - x + 1 = 16 2 x 2 - x - 15 = 0 (2 x + 5)( x - 3) = 0 The original price of an apple is \$3. 2x + 5 = 0 11. Let x be the number of articles that the man bought, then original cost of one article is \$ 300 . x x = - or 5 (rejected) or 2 x-3= 0 x = 3 14. Q Area of ABCD -area ofABP -area ofADQ - area of CPQ = shaded area x 300 15 - 300 = 25 4 x 15 x 7500 - 300 = 4 x 15 x 2 - 1200 x = 30 000 x 2 - 80 x - 2000 = 0 ( x - 100)( x + 20) = 0 22 Certificate Mathematics in Action Full Solutions 4A 72 - 7( 7 - x ) 7( 7 - 2 x ) 2 x x - - = 17 2 2 2 49 - 7 x 49 - 14 x 2x2 49 - - - = 17 2 2 2 98 - 49 + 7 x - 49 + 14 x - 2 x 2 = 34 2 x 2 - 21x + 34 = 0 ( 2 x - 17)( x - 2) = 0 2 x - 17 = 0 or x - 2 = 0 x = 2 x = 8.5 (rejected) or DC = EC 2 + DE 2 (Pyth. theorem) = 62 + 82 cm = 10 cm Perimeter of trapezium ABCD = (8 + 6.5 + 10 + 12.5) cm = 37 cm 15. (a) Area of trapezium ABCD [( x + 2) + (3 x - 1)]( 2 x - 1) cm 2 2 ( 4 x + 1)( 2 x - 1) 2 = cm 2 = 8x - 2 x - 1 2 = cm 2 1 2 2 = 4 x - x - cm 2 2 Revision Exercise 1 (p. 44) Level 1 1. 6 x 2 + 11x - 10 = 0 (3x - 2)( 2 x + 5) = 0 3x - 2 = 0 or 2 x + 5 = 0 2 5 x = or x = - 3 2 (b) (i) By (a), 1 = 76 2 8 x 2 - 2 x - 1 = 152 4x2 - x - 8 x 2 - 2 x - 153 = 0 2. 12 x 2 - 3 = 35 x 12 x 2 - 35 x - 3 = 0 (12 x + 1)( x - 3) = 0 12 x + 1 = 0 x = - or 1 or 12 x-3= 0 x = 3 (ii) 8 x 2 - 2 x - 153 = 0 ( 4 x + 17)( 2 x - 9) = 0 4 x + 17 = 0 or 2 x - 9 = 0 17 9 x = - (rejected) or x = 4 2 3. Q x = -4 or x = 6 x + 4 = 0 or x - 6 = 0 ( x + 4)( x - 6) = 0 x 2 - 2 x - 24 = 0 The required equation is x2 - 2x - 24 = 0. EC = BC - BE = BC - AD 9 9 = 3 - 1 - + 2 cm 2 2 = (12.5 - 6.5) cm = 6 cm 4. 1 3 or x = 3 2 3 1 x- = 0 or x - = 0 2 3 2 x - 3 = 0 or 3x - 1 = 0 x = ( 2 x - 3)( 3x - 1) = 0 6 x 2 - 11x + 3 = 0 23 1 The required equation is 6x2 -11x + 3 = 0. 2 Quadratic Equations in One Unknown 5. Using the quadratic formula, 1 = - 4( -3)(1) 4 193 = 16 > 0 The equation - 3x 2 + distinct real roots. 11. Q The equation x2 -4x + 2k = 0 has a double real root. = 0 ( -4) 2 - 4(1)( 2k ) = 0 16 - 8k = 0 k = 2 The equation kx2 + 3x -6 = 0 has a double real root. = 0 (3) 2 - 4(k )( -6) = 0 9 + 24k = 0 k = - 3 8 1 x + 1 = 0 has two 4 x= = = - b b 2 - 4ac 2 - (-3) (-3) - 4(4)(-4) 2( 4) 2 3 73 8 3 + 73 3 - 73 = or 8 8 = 1.44 (cor. to 2 d.p.) or - 0.69 (cor. to 2 d.p.) 6. Using the quadratic formula, - b b 2 - 4ac x= 2 = - (-3) ( -3) - 4(7)(-5) 2(7 ) 2 12. Q 3 149 = 14 3 + 149 3 - 149 = or 14 14 = 1.09 (cor. to 2 d.p.) or - 0.66 (cor. to 2 d.p.) 13. Q The equation x2 + 6x + k = 0 has no real roots. <0 7. For 2x2 - 5x + 1 = 0 = ( -5) - 4( 2)(1) = 17 2 14. Q 6 2 - 4(1)(k ) < 0 36 - 4k < 0 k >9 The range of possible values of k is k > 9. The equation 3kx2 -4x + 3 = 0 has no real roots. <0 >0 The equation 2x2 - 5x + 1 = 0 has two distinct real roots. (-4) 2 - 4(3k )(3) < 0 16 - 36k < 0 4 k> 9 The range of possible values of k is k > - 1 8 0 3 1 0 2 - 1 3 0 4 3 5 8 4 . 9 8. For x2 - 8x + 16 = 0 = ( -8) 2 - 4(1)(16) = 0 = 0 9. The equation x2 - 8x + 16 = 0 has a double real root. 15. (a) x y For 2x2 + 3x + 4 = 0 = (3) 2 - 4( 2)( 4) = - 23 < 0 The equation 2x + 3x + 4 = 0 has no real roots. 1 x +1= 0 4 2 (b) 10. For - 3x 2 + 24 Certificate Mathematics in Action Full Solutions 4A Alternative Solution Let x cm be the length of the rectangle, then the width of the rectangle. 88 2 x + = 38 x 88 x+ = 19 x x 2 + 88 = 19 x x 2 - 19 x + 88 = 0 ( x - 11)( x - 8) = 0 x - 11 = 0 or x - 8 = 0 x = 11 or x = 8 The length and width of the rectangle are 11 cm and 8 cm respectively. 8x - 4 x 2 = 3 4x 2 - 8x + 3 = 0 ( 2 x - 1)( 2 x - 3) = 0 2 x - 1 = 0 or 2 x - 3 = 0 x = 0.5 or x = 1.5 After 0.5 seconds and 1.5 seconds, the ball is 3 m above the ground. The equation x2 -ax -40 = 0 has two distinct real roots. > 0 ( -a ) 2 - 4(1)( -40) > 0 a 2 + 160 > 0 e a 2 > -160 The square of any numbers is always positive. The equation x2 -ax -40 = 0 has two distinct real roots for any real values of a. a = 1 or 2 or 3 (or any other reasonable answers) 2 88 cm is x 16. (a) two (b) The x-intercepts of y = 4x2 -4x -3 are - and 1.5. 0.5 Therefore, the roots of the equation 4x2 - 4x - 3 = 0 are - and 1.5. 0.5 17. (a) one (b) The x-intercepts of y = x2 -6x + 9 is 3. Therefore, the roots of the equation x2 - 6x = - is 3. 9 18. Let x be one of the number, then 27 -x is the other number. x ( 27 - x ) = 180 2 20. 21. Q 27 x - x 2 = 180 x - 27 x + 180 = 0 ( x - 15)( x - 12) = 0 x - 15 = 0 or x - 12 = 0 x = 15 or x = 12 The two numbers are 12 and 15. 180 is the other x Alternative Solution Let x be one of the number, then number. 180 = 27 x x 2 + 180 = 27 x x+ x 2 - 27 x + 180 = 0 ( x - 15)( x - 12) = 0 x - 15 = 0 or x - 12 = 0 x = 15 or x = 12 The two numbers are 12 and 15. 2x - 1 = m 3 22. 4 1 4x 2 - x + - m = 0 3 9 Using the quadratic formula, x = -b b 2 - 4ac 2a 4 1 - - 4( 4 ) - m 3 9 2( 4) 2 19. Let x cm be the length of the rectangle, then 38 - 2 x = (19 - x ) cm is the width of the rectangle. 2 x (19 - x ) = 88 2 4 - - 3 = 19 x - x 2 = 88 x - 19 x + 88 = 0 ( x - 11)( x - 8) = 0 x - 11 = 0 or x - 8 = 0 x = 11 or x = 8 The length and width of the rectangle are 11 cm and 8 cm respectively. 4 16 m = 3 8 To have two rational roots of different signs, we need 4 16 m > 3 16 16 m > 9 1 m > 9 m = 1 or 9 or 16 (or any other reasonable answers) 23. Q The graph of y = ax2 + 4x + c intersects the x-axis at 25 1 one point. The equation ax2 + 4x + c = 0 has a double real root. = 0 ( 4) 2 - 4ac = 0 16 - 4ac = 0 ac = 4 a = - c = - or a = 1, c = 4 or a = 2, c = 2. 2, 2 (or any other reasonable answers) The graph y = - + 2x + k does not intersect the x x-axis. The equation - 2 + 2x + k = 0 has no real roots. x < 0 2 Quadratic Equations in One Unknown 2( x + 1)( x - 1) = 5 x + 4 2( x 2 - 1) = 5 x + 4 2 x2 - 2 = 5x + 4 2 x2 - 5x = 6 5 x2 - x = 3 2 x2 - 27. 5 5 5 x + = 3+ 2 4 4 x- 5 73 = 4 16 5 73 = 4 4 x= = 5 73 4 4 5 + 73 4 or 5 - 73 4 2 2 2 24. Q x- 2 2 - 4( -1)( k ) < 0 4 + 4k < 0 k < -1 k = - or - or - (or any other reasonable answers) 4 3 2 Level 2 x( x - 6) + 4 = 0 x2 - 6x + 4 = 0 x 2 - 6 x = -4 25. 6 6 x 2 - 6 x + = -4 + 2 2 2 ( x - 3) = 5 x-3 = 5 x = 3 5 = 3+ 5 16( x 2 + 4) = 64 x + 9 16 x 2 + 64 = 64 x + 9 16 x 2 - 4 x = -55 55 x2 - 4x = - 16 26. 55 4 4 x2 - 4 x + = - + 16 2 2 9 ( x - 2) 2 = 16 3 x-2= 4 3 x = 2 4 11 5 = or 4 4 2 2 2 2 (2 x + 3)( x - 4) = 3x - 2 2 x 2 - 5 x - 12 = 3x - 2 2 x 2 - 8 x = 10 x2 - 4 x = 5 28. or 3- 5 4 4 x2 - 4x + = 5 + 2 2 2 ( x - 2) = 9 x - 2 = 3 x = 23 x = 5 or - 1 x (6 x - 5) = 4 6x2 - 5x = 4 6 x2 - 5x - 4 = 0 (3 x - 4)(2 x + 1) = 0 3x - 4 = 0 4 x= 3 2x 3 9( x 2 - 1) = 2 x 3( x 2 - 1) = 9x2 - 9 = 2x 9x - 2x - 9 = 0 Using the quadratic formula, 2 2 2 29. or or 2x + 1 = 0 x =- 1 2 30. 26 Certificate Mathematics in Action Full Solutions 4A - b b 2 - 4ac 2a - ( -2) (-2) 2 - 4(9)(-9) 2(9) 3 (b) The roots of the required equation are 2 and 2 2(4), i.e. 3 and 8. e or 1 - 82 9 34. (a) x = 3 or x=8 x - 3 = 0 or x - 8 = 0 ( x - 3)( x - 8) = 0 x 2 - 11x + 24 = 0 The required equation is x 2 - 11x + 24 = 0 . x= = 2 328 = 18 = 2 82 18 1 + 82 = 9 3x 2 + 8 x - 3 = 0 (3 x - 1)( x + 3) = 0 3x - 1 = 0 1 x= 3 or or x+3= 0 x = -3 ( 2 x + 1)(3 x - 1) = 4( x - 1)(2 x + 1) 31. 6 x2 + x - 1 = 8x2 - 4x - 4 2 x2 - 5x - 3 = 0 (2 x + 1)( x - 3) = 0 2x + 1 = 0 or x - 3 = 0 x=- 1 2 or x=3 Alternative Solution ( 2 x + 1)(3 x - 1) = 4( x - 1)(2 x + 1) ( 2 x + 1)(3x - 1) - 4( x - 1)(2 x + 1) = 0 ( 2 x + 1)[(3 x - 1) - 4( x - 1)] = 0 ( 2 x + 1)(- x + 3) = 0 ( 2 x + 1)( x - 3) = 0 2x + 1 = 0 1 x=- 2 or or x-3 = 0 x=3 1 1 (b) The roots of the required equation are 1 and . -3 3 1 i.e. 3 and - . 3 e 35. (a) x=3 x-3 = 0 or or x=- x+ 1 3 1 =0 3 x - 3 = 0 or 3 x + 1 = 0 ( x - 3)(3 x - 1) = 0 3x 2 - 8 x - 3 = 0 The required equation is 3 x 2 - 8 x - 3 = 0 . ( x - 1) 2 + 4( x - 1) + 4 = 0 x - 2x + 1 + 4x - 4 + 4 = 0 2 32. x2 + 2x + 1 = 0 ( x + 1) 2 = 0 x +1 = 0 x = -1 Alternative Solution Q The equation x 2 - 2( k - 5) x + 16 = 0 has a double real root. =0 [-2( k - 5)]2 - 4(1)(16) = 0 4( k - 5) 2 - 64 = 0 (k - 5) 2 - 16 = 0 (k - 5) 2 = 16 k - 5 = 4 k = 5 4 =9 or ( x - 1) 2 + 4( x - 1) + 4 = 0 [ ( x - 1) + 2] 2 = 0 ( x + 1) 2 = 0 x +1 = 0 x = -1 33. (a) 1 2 x 2 - 11x + 12 = 0 (2 x - 3)( x - 4) = 0 2x - 3 = 0 3 x= 2 or or x-4 = 0 x=4 (b) For k = 9 , x 2 - 2(9 - 5) x + 16 = 0 x 2 - 8 x + 16 = 0 ( x - 4) 2 = 0 x-4= 0 x=4 For k = 1 , 27 1 x 2 - 2(1 - 5) x + 16 = 0 x 2 + 8 x + 16 = 0 ( x + 4) 2 = 0 x+4=0 x = -4 36. (a) Q The equation x 2 - 3x - ( k + 2) = 0 has real roots. 0 ( -3) - 4(1)[-( k + 2)] 0 9 + 4(k + 2) 0 4k + 17 0 2 Quadratic Equations in One Unknown 8 x 2 - 8 x + 15 = 0 15 =0 (ii) 4 x 2 - 4 x + 2 19 4x2 - 4x + =2 2 19 >3 e 2 The equation has no real roots. 38. (a) x y - 3 16 - 2 5 - 1 - 2 0 - 5 1 - 4 2 1 3 10 k- 17 4 17 . 4 (b) The range of possible values of k is k - 17 4 17 . 4 (b) Q k- Minimum value of k is - 17 x 2 - 3x - - + 2 = 0 4 9 2 x - 3x + = 0 4 (c) 3 x - = 0 2 3 x- =0 2 3 x= 2 4 x2 - 4 x + k = 2 4 x - 4 x + (k - 2) = 0 2 2 (c) The x-intercepts of y = 2 x 2 - x - 5 read from the graph are - and 1.9. 1.4 Therefore, the roots of the equation 2 x 2 - x - 5 = 0 are - and 1.9. 1.4 39. (a) x y (b) - 2 10 - 1 3 0 - 2 1 - 5 2 - 6 3 - 5 4 - 2 5 3 37. (a) e The equation 4 x 2 - 4 x + ( k - 2) = 0 has no real roots. <0 ( -4) 2 - 4( 4)(k - 2) < 0 16 - 16( k - 2) < 0 - 16k + 48 < 0 k >3 The range of possible values of k is k > 3. 4 x2 - 4 x - 1 = 0 4x2 - 4x + 1 = 2 e 1<3 The equation has real roots. 40. (a) x y 1 9 2 4 3 1 4 0 5 1 6 4 7 9 (b) (i) (c) The x-intercepts of y = x 2 - 4 x - 2 read from the graph are - and 4.4. 0.4 Therefore, the roots of the equation x 2 = 2( 2 x + 1) are - and 4.4. 0.4 28 Certificate Mathematics in Action Full Solutions 4A 43. Let x km/h be the original speed of the car. 240 240 - = x x + 15 240( x + 15 - x) = x( x + 15) 5 (240)(15) = 4 4500 = 2 4 5 4 5 x( x + 15) x 2 + 15 x (b) (i) 1 double real root x + 15 x - 4500 = 0 ( x - 60)( x + 75) = 0 x - 60 = 0 or x + 75 = 0 x = 60 or x = - 75 (rejected) The original speed of the car is 60 km/h. 44. Let V m3 be the volume of the swimming pool, and x h be the time the larger pipe take to fill up the swimming pool, then ( x + 6) h is the time the smaller pipe take to fill up the swimming pool and V 3 V + m is the volume of water filled in the x x+6 swimming pool in 1 hour using two pipes. V =4 V V + x x+6 1 1 V = 4V + x x+6 x + 6+ x 1 = 4 x( x + 6) x 2 + 6 x = 8 x + 24 x 2 - 2 x - 24 = 0 ( x - 6)( x + 4) = 0 (ii) The x-intercept of y = x 2 - 8 x + 16 is 4. Therefore, the root of the equation x(8 - x) = 16 is 4. 41. Q The graph of y = kx + 5 x + k touches the x-axis.. The equation kx 2 + 5 x + k = 0 has a double real root. 2 =0 5 - 4(k )(k ) = 0 2 25 - 4k 2 = 0 (5 + 2k )(5 - 2k ) = 0 5 + 2k = 0 or 5 - 2k = 0 5 5 k=- or k= 2 2 42. Q The graph of y = -2 x 2 + 3 x - k - 1 has two distinct x-intercepts. The equation - 2 x 2 + 3 x - k - 1 = 0 has two distinct real roots. >0 ( 3 )2 - 4( - 2 )( - k - 1 ) > 0 9 - 8(k + 1 ) > 0 1 - 8k > 0 1 k< 8 The range of possible values of k is k < 1 . 8 x-6 =0 x=6 or or x+4=0 x = -4 (rejected) It would take the larger pipe 6 hours to fill up the swimming pool. AC 2 + BC 2 = AB 2 (Pyth. theorem) 45. (a) ( x + 3) 2 + (7 x - 2) 2 = (6 x + 5) 2 x 2 + 6 x + 9 + 49 x 2 - 28 x + 4 = 36 x 2 + 60 x + 25 14 x 2 - 82 x - 12 = 0 7 x 2 - 41x - 6 = 0 7 x 2 - 41x - 6 = 0 (7 x + 1)( x - 6) = 0 7x + 1 = 0 or 1 x = - (rejected) or 7 (b) x-6 = 0 x=6 29 1 AC BC 2 (6 + 3) (7 6 - 2) = cm 2 Area of ABC 2 9 40 = cm 2 2 = 180 cm 2 = Perimeter of ABC = AB + BC + AC = [(6 6 + 5) + (7 6 - 2) + (6 + 3)] cm = ( 41 + 40 + 9) cm = 90 cm Quadratic Equations in One Unknown 4. x = -5 or x=2 x + 5 = 0 or x - 2 = 0 ( x + 5)( x - 2) = 0 x 2 + 3x - 10 = 0 The required equation is y = x 2 + 3x - 10 . Answer: D For x 2 + 2 x + 3 = 0 , = 22 - 4(1)(3) = -8 <0 The equation x 2 + 2 x + 3 = 0 has no real roots. The graph y = x 2 + 2 x + 3 has no x-intercepts. For x = 0 y = 0 2 + 2(0) + 3 =3 The graph y = x 2 + 2 x + 3 has positive y-intercept. 46. (a) Area of shaded region 43 = (5 x - 2)(3 x + 1) - 8 cm 2 2 = (15 x 2 - x - 2 - 48) cm 2 = (15 x 2 - x - 50) cm 2 (b) (i) Q 15 x 2 - x - 50 = 186 5. Answer: B Q The graph y = x 2 - 4 x + c touches the x-axis. The equation x 2 - 4 x + c = 0 has a double real root. =0 ( -4) 2 - 4(1)(c) = 0 16 - 4c = 0 c=4 15 x 2 - x - 236 = 0 (ii) 15 x 2 - x - 236 = 0 (15 x + 59)( x - 4) = 0 15 x + 59 = 0 or x - 4 = 0 59 x=- (rejected) or x=4 15 JK = BC - BJ - KC = [(3 4 + 1) - 4 - 4] cm = 5 cm GH = AB - AG - HB = [(5 4 - 2) - 3 - 3] cm = 12 cm 6. Answer: A Q The equation x 2 + ( k + 8) x + 8k = 0 has a double root. =0 (k + 8) - 4(1)(8k ) = 0 2 k 2 + 16k + 64 - 32k = 0 k 2 - 16k + 64 = 0 (k - 8) 2 = 0 k -8 = 0 k =8 Multiple Choice Questions (p. 48) 1. Answer: C 2x2 - x - 6 = 0 ( 2 x + 3)( x - 2) = 0 2x + 3 = 0 or x - 2 = 0 x=- 3 or 2 7. Answer: A Q The equation x 2 - 8 x + p = 0 has no real roots. <0 ( -8) 2 - 4(1)( p ) < 0 64 - 4 p < 0 p > 16 The range of possible values of p is p > 16. x=2 2. Answer: B 16 x 2 + 8 x + 1 = ( 4 x + 1) 2 3. Answer: D Q The x-intercepts of the graph are 5 and 2. The roots of the graph are 5 and 2. 8. Answer: D ( x - 2)( x - 4) = ( x - 2) ( x - 2)( x - 4) - ( x - 2) = 0 ( x - 2)[( x - 4) - 1] = 0 ( x - 2)( x - 5) = 0 30 Certificate Mathematics in Action Full Solutions 4A x - 2 = 0 or x - 5 = 0 x = 2 or x=5 9. Answer: B Q is a root of 2 x 2 + 3 x - 5 = 0 . 2 + 3 - 5 = 0 4 + 6 - 5 = 2(2 + 3 - 5) + 5 = 2(0) + 5 . =5 2 2 2 2 5 1 - - - - 4 - 3 3 2 4 x= 5 2 2 2 53 3 18 = 5 = = 2 106 15 30 4 + 106 30 or 4 - 106 30 2 10. Answer: A Q The graph y = x 2 - 8 x + c has two x-intercepts. The equation x 2 - 8 x + c = 0 has two distinct real roots. >0 ( -8) 2 - 4(1)(c) > 0 64 - 4c > 0 c < 16 Ken's method: 5 2 2x 1 x - = 2 3 4 30 x 2 - 8 x = 3 30 x 2 - 8 x - 3 = 0 Using the quadratic formula, The range of possible values of c is c < 16. x= = = = = - ( -8) (-8) 2 - 4(30)(-3) 2(30) 8 424 60 8 2 106 60 4 106 30 4 + 106 30 or 4 - 106 30 HKMO (p. 49) f ( x) + kg ( x) = 0 41x - 4 x + 4 + k ( -2 x + x) = 0 2 2 ( 41 - 2k ) x 2 - ( 4 - k ) x + 4 = 0 e The equation f ( x ) + kg ( x) = 0 has a single root. =0 [-(4 - k )]2 - 4( 41 - 2k )(4) = 0 k 2 - 8k + 16 - 656 + 32k = 0 k 2 + 24k - 640 = 0 (k - 16)(k + 40) = 0 k - 16 = 0 or k + 40 = 0 or k = -40 k = 16 d = -40 Let's Discuss p. 20 Angel's method: 5 2 2x 1 x - = 2 3 4 5 2 2x 1 x - - =0 2 3 4 Using the quadratic formula, 31 1 Quadratic Equations in One Unknown 32 ...
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## This document was uploaded on 09/23/2009.

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