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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 xintercepts 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 Followup 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 x3= 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) 32 1 2 p. 16 x2  8x  9 = 0 x2  8x = 9
1. (a) 2. (a) ( x  2) 2 = 5 x2 = 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 x3= 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 xintercepts 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 xintercept 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 x4= 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. x3= 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 xd = 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 x4 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 x8 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 xintercepts 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 xintercepts 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 xintercepts 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 xintercepts 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 xintercepts 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 xintercept 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 xaxis. (b) x2 = x + 4 x  x4 = 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 xintercepts 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 xaxis 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 xaxis. 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 xintercept 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 xintercept 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 xaxis. 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 xaxis. 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 x6 = 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 x4 = 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 x3= 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 x3= 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 xintercepts 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 xintercepts 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 xaxis 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 xaxis. 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 x3 = 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 x2= 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 x3 = 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 x3 = 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 x4 = 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 x4= 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 xintercepts 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 xintercepts 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 xintercept 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 xaxis.. 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 xintercepts. 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 x6 =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) x6 = 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 xintercepts. For x = 0 y = 0 2 + 2(0) + 3 =3 The graph y = x 2 + 2 x + 3 has positive yintercept. 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 xaxis. 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 xintercepts 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 xintercepts. 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.
 Spring '09
 Math, Equations

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