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

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

This document was uploaded on 09/23/2009.

Ask a homework question - tutors are online