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F.4 Maths - Variation

發問:

y varies jointly as x+p & x+3, where p is a constant. When x=2, y=70. Whenx=3, y=96. (a) find the value of p (b)find an equation connecting x and y (c)find the values of x when y=160

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最佳解答:

(a)Let y = k(x+p)(x+3), where k is a non-zero constant From the given, we get 70 = k(2+p)(2+3) k(2+p) = 14----------(1) 96 = k(3+p)(3+3) k(3+p) = 16-----------(2) (1)/(2), (2+p)/(3+p) = 14/16 16(2+p) = 14(3+p) p = 5 (b)Substitute p = 5 into (1), we get k = 2 Therefore, the equation connectiong x and y is y = 2 (x+5) (x+3) (c) When y = 160, 2 (x+5) (x+3) = 160 x^2+8x+15 = 80 x^2+8x-65 = 0 (x-5) (x+13) = 0 x = 5 or -13 希望幫到你

其他解答:

Let y = k(x+p)(x+3) = k[x^2 + (p+3)x +3p] when x = 2, y = 70 70 = k ( 4 + 2p + 6 + 3p ) 70 = k ( 5p + 10 ) 14 = k ( p + 2 ) ................... [1] when x = 3, y = 96 96 = k ( 9 + 3p + 9 + 3p ) 96 = k ( 6p + 18 ) 16 = k ( p + 3 ) ...................... [2] [2] - [1], k = 2 p = 16/2 - 3 = 5 Thus, y = 2 [x^2 + (5+3)x +3(5)] = 2 ( x^2 + 8x + 15) when y = 160, x^2 + 8x + 15 = 80 x^2 + 8x - 65 = 0 (x + 13) ( x - 5) = 0 x = -13 or 5