F.4 Quadratic equation

[匿名]

2009-12-26 8:28 pm

1)
The area and the perimeter of a rectangle are 126 cm^2
and 46 cm respectively.Find the difference between the
length and the width of the rectangle.

2)
In a two-digit number,the sum of the digits is 9 and the
square of the tens digit is smaller than twice the units digit
by 3.Find the two-digit number.

3)The length of a rectangle is 4 cm longer than twice its width.
If its area is 70 cm^2,find the length.

Please show clear steps.Thanks.

回答 (1)

Prof. Physics

2009-12-26 8:40 pm

✔ 最佳答案

1. Let x and y be the length and width of the rectangle respectively.

xy = 126 ... (1)

2(x + y) = 46 ... (2)

From (2): x + y = 23

(x + y)^2 = 529

(x - y)^2 + 4xy = 529

(x - y)^2 + 4(126) = 529

(x - y)^2 = 25

x - y = 5 (just need to consider the positive root)

So, the difference is 5.

2. Let the tens digit be x, the units digit be y

x + y = 9 ... (1)

x^2 = 2y + 3 ... (2)

Sub (1) into (2):

x^2 = 2(9 - x) + 3

x^2 + 2x - 15 = 0

(x - 3)(x + 5) = 0

x = 3 or -5 (rejected)

So, y = 6

Therefore, the required number is 36.

3. Let x and y be the length and width respectively.

x = y + 4 ... (1)

xy = 70

x(x - 4) = 70

x^2 - 4x - 70 = 0

x = {-(-4) +- sqrt((-4)^2 - 4(1)(-70))}/2

= 2 +- sqrt74

We need to postive root

So, the length = 2 + sqrt74 cm

參考: Physics king

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