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There are three water pipes A,B and C each with water flowing at a constant speed. If the pipes are used individually to fill up an empty pool, pipe A takes 5 more hours to fill it up than pipe B, while pipe C takes 2 more hours than pipe B. If the three pipes are used simultaneously, they take 4 hours to fill up the pool. Suppose pipe B alone takes x hours to fill up the pool.
a) Prove that x^3 - 5x^2 - 46x - 40 = 0
b) Hence find x by factorizing x^3 - 5x^2 - 46x - 40
S4 polynomial
2012-03-18 12:40 pm
回答 (3)
2012-03-18 2:42 pm
✔ 最佳答案
a)Suppose pipe B alone takes x hours to fill up the pool.
Then , pipe A and pipe C takes ( x + 5 ) and ( x + 2 ) hours to fill up the pool respectively.Let the volume of the pool be 1 cubic unit.
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參考: 自己
2012-03-19 4:27 am
both good enough, just Jan Man shows more steps that is more suitable for the foolish like me
2012-03-18 6:41 pm
(a) The speed of pipe B is 1/x pool/hr
So, the speeds of pipes A and C are 1/(x + 5) and 1/(x + 2) respectively
Now, they take 4 hours to fill up the pool.
So, 4[1/x + 1/(x + 5) + 1/(x + 2)] = 1
4(x + 2)(x + 5) + 4x(x + 2) + 4x(x + 5) = x(x + 2)(x + 5)
4(x^2 + 7x + 10 + x^2 + 2x + x^2 + 5x) = x^3 + 7x^2 + 10x
4(3x^2 + 14x + 10) = x^3 + 7x^2 + 10x
x^3 - 5x^2 - 46x - 40 = 0
(b) f(x) = x^3 - 5x^2 - 46x - 40
f(-1) = -1 - 5 + 46 - 40 = 0
So, f(x) = (x + 1)(x^2 - 6x - 40) = (x + 1)(x - 10)(x + 4)
x = 10 (since x is positive)
So, the speeds of pipes A and C are 1/(x + 5) and 1/(x + 2) respectively
Now, they take 4 hours to fill up the pool.
So, 4[1/x + 1/(x + 5) + 1/(x + 2)] = 1
4(x + 2)(x + 5) + 4x(x + 2) + 4x(x + 5) = x(x + 2)(x + 5)
4(x^2 + 7x + 10 + x^2 + 2x + x^2 + 5x) = x^3 + 7x^2 + 10x
4(3x^2 + 14x + 10) = x^3 + 7x^2 + 10x
x^3 - 5x^2 - 46x - 40 = 0
(b) f(x) = x^3 - 5x^2 - 46x - 40
f(-1) = -1 - 5 + 46 - 40 = 0
So, f(x) = (x + 1)(x^2 - 6x - 40) = (x + 1)(x - 10)(x + 4)
x = 10 (since x is positive)
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