F.4 Quadratic Equation

2009-09-24 4:41 am

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2009-09-24 7:36 am
✔ 最佳答案
(a) Given the graph y = x^2 - 20x + 40
x^2 - 20x + 40 = 0 => y = 0.
The solution can be obtained by referring to the points where the curve cuts the x-axis.
The values of x at these 2 points are approximately 2.3 and 17.7
(b) (i) T = 0.25x^2 - 5x + 30
When the temperature is 20, 20 = 0.25x^2 - 5x + 30
0.25x^2 - 5x + 10 = 0
x^2 - 20x + 40 = 0 which is exactly the same equation as in (a). So graphically the solution is x = 2.3
At x = 2.3, the time is 8.3 hours after noon, or 8:18pm approximately
(ii) Actually the result obtained in (bi) is not very accurate. Since the smallest scale in the graph is 1 unit, the absolute error is up to 0.5 unit which is half an hour. I would prefer to use the quadratic equation formula to obtain the result as follows:
x^2 - 20x + 40 = 0
x = [20 +/- √(400 - 160)] / 2 = 2.254 taking the smaller root. This is equivalent to 8:15pm


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