J17

21. 2y 2 1
– 5y + 1 = 0 22. y 2 +3y +1= 0
23. 2 – x – 6x 2 = 0 24. 3 + 4x – 2x 2 = 0
25. 1 – 5x – 2x 2 = 0 26. 3x 2 – 1 + 4x = 0
27. 5x – x 2 + 2 = 0 28. 24x 2 – 22x – 35 = 0
29. 36x 2 – 17x – 35 = 0 30. 20x 2 + 17x – 63 = 0
31. x 2 + 2.5x – 6 = 0 32. 0.3y 2 + 0.4y – 1.5 = 0
33. 10 – x – 3x 2 = 0 34. x 2 + 3.3x – 0.7 = 0
35. x 2 = 6 – x 36. x(x + 10) = -21
37. 3x + 2 = 2x 2 38. x 2 + 4 = 5x
39. 6x(x + 1) = 5 – x 40. (2x – 1) 2 = (x – 1) 2 + 8
7 2 2
41. 2x + 2 = - 1 42. + = 3
x x x +1
43. 3x(x+2) – x(x-2) + 6 = 0 44. (x – 3) 2 = 10
Problems solved by quadratic equations
Example
The perimeter of a rectangle is 42 cm. If the diagonal is 15 cm, find the width of the
rectangle.
Solution
Let the width of the rectangle be x cm.
Since the perimeter is 42 cm, the sum of the length and the width is 21 cm.
Therefore, length of rectangle = (21 – x) cm.
2
By Pythagoras’ theorem
x 2 + (21 – x)2 = 152
x 2 + 441 – 42x + x 2 = 225
2x 2 – 42x + 216 = 0
x 2 – 21x + 108 = 0
(x – 12)(x – 9) = 0
x = 12 or x = 9
note that the dimensions of the rectangle are 9 cm by 12 cm, whichever value of x is
taken.
Therefore, the width of the rectangle is 9 cm.