By J. Newbury PhD (auth.)
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V=wya 2 -x 2 ;x W 2/ 2 = L - / . -LT-lt' 33. L V = y [gd(1 + ~)J 35. C = 37. J(a2 + b2 ) ;a = -2W (R 2 - r2 ) " r a + 2b h + b) ; X = 3(a a 1 1 u v 26. - + - 2 = -f ; u 28. A - y'-2P2 _ Q2 29. A = 1T,y(h 2 + r 2 + 1Tr 2 ):h . d _ (p 2 _ 2Q2) . S3\ %y (g;) ; 8 31 e - y(Z); H 24. A=1Tr 2 y(h 2 +r 2 );h 25. T = H + - - ; (i) H 4 (ii) 1 27. M="4 (1-"2);1 19 H = y(3;); h 30. r = 32. E ; d ,P ~ + y(~2 + q2} = 2Hg - v2 • Wf ' ; q H 34. T = 21Ty (}12g ; k2) 36. E +~)= 3k(1-~);m 38. W 32x ' 2 5 Quadratic Expressions The type of equation I have so far dealt with is usually known as a 'linear' equation because of its connection with a straight line graph.
P -1 E 3 G • -3 -4 -5 L Fourth quadrant Figure 4 For the next exercise you will need to draw your own axis system on graph paper; this is a straightforward task and ü you wish you can simply copy the system we have drawn for Exercise 121, although you do not have to use the scale we have used. You can choose whatever scale you wish. It is usual to use the same scale for both the x and y axes although the finished product would then look rather different to figure 4 above. Graphs and Co-ordinates Exercise 122: Exercise 123: 35 Draw an axis system and mark on the following points: A (4,2), B (5,0), C (0,1), D (-2, 4), E (3, -1), F (--4, -5), G (0, -2), H (-5,0).
If x = 1 is true that x + 2x-3 = O? And if x = -3? In other words, are these values of x the values which fit the equation x 2 + 2x - 3 = O? So we can see from this that if we can express a quadratic expression like x 2 + 2x - 3 = 0 in bracket form, then we can certainly find two values which x can take and which fit the equation. In other words, we can solve the equation. Exercise 112: Exercise 113: Exercise 114: Exercise 115: Express x 2 -7x + 12 in bracket form. Solve x 2 - 7x + 12 = 0 by repeating the sequence followed in Exercises 106 -111.