Quadratic Equation

1.
Ifthe length and the width of a rectangle are 1cm and 8cm less than its diagonalrespectively, findthe perimeter of the rectangle.

Let the d cm be the diagonal of the rectangle.
Then, length of the rectangle = (d - 1) cm
and width of the rectangle = (d - 8) cm

By Pythagorean theorem:
d² = (d - 1)² + (d - 8)²
d² = d² - 2d + 1 + d² - 16d + 64
d² - 18d + 65 = 0
(d - 5)(d - 13) = 0
d = 5 (rejected) or d = 13

Hence, the perimeter of the rectangle
= 2[(13 - 1) + (13 - 8)] cm
= 34 cm


= = = = =
2.11.
At present, theage of Anthony is three times the age of his sister. Fiveyears later, thedifference between the squares of their ages is 300. Findthe present ages of Anthony and his sister.

Let s be the age of Anthony's sister.
Then, the age of Anthony = 3s

(3s + 5)² - (s + 5)² = 300
9s² + 30s + 25 - s² - 10s - 25 = 300
8s² + 20s - 300 = 0
2s² + 5s - 75 = 0
(2s + 15)(s - 5) = 0
s = -15/2 (rejected) or s = 5
3s = 15

Present age of Anthony = 15
Present age of his sister = 5


= = = = =
3.
Thevalue of a computer is $10000 now and its value is $7200after two years. If its annual depreciation are rate of thesecond year is two times that of the first year, findthe depreciation rate of the first year.

Let the annual depreciation rate of the first year = p%
Then, the annual depreciation rate of the second year = 2p%

10000 ´ (1- p%) ´ (1- 2p%) = 7200
25 [1 - 3p% + 2(p%)²] = 18
25 - 75p% + 50(p%)² = 18
50(p%)² - 75(p%) + 7 = 0
(10p% - 1)(5p% - 7) = 0
p% = 1/10 or p% = 7/5 (rejected)
p% = 10%

The annual depreciation rate of the first year = 10%

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2Ans:
Let X is the age of Anthony' sister.
So 3X is the age of Anthony.
(3X+5)2-(X+5)=300
8X2+20X-300=0
X=5 or -7.5(rej.)
¡ïthe age of Anthony' sister is 5.
3(5)=15
¡ïthe age of Anthony is 15.

2011-10-25 23:02:24 補充：
3.Ans:
Let X is the depreciation rate
So 2X isthe depreciation rate .
10000(1-X)(1-2X)=7200
1-3X+2X2=72/100
2X2-3X+7/25=0
x=1.4(rej.) or 0.1=10%
¡ïtthe depreciation rate is 10%