maths problems

1)

x^2 + 15^2 = (3x - 7)^2

x^2 + 225 = 9x^2 - 42x + 49

8x^2 - 42x - 176 = 0

4x^2 - 21x - 88 = 0

(4x + 11)(x - 8) = 0

x = - 11/4 (rejected)

or x = 8

The required length = 8 cm




圖片參考：http://img69.imageshack.us/img69/9246/maths3.png



2a)
2(AD + x) = 20

AD = 10 - x

Area = AD * x

a = (10 - x)x

a = 10x - x^2


2b)

Taking a = 20 ,

20 = 10x - x^2

x^2 - 10x + 20 = 0

x = [ -(-10) +/- √(10^2 - 4(1)(20)) ] / 2

x = (5 +/- √5)

x = 7.236.... = 7.2 (2 dec.)
or
x = 2.7639... = 2.8 (2 dec.)

Taking a = 18 ,

18 = 10x - x^2

x^2 - 10x + 18 = 0

x = [- (-10) +/- √(10^2 - 4(1)(18))] / 2

x = 5 +/- √7

x = 7.64... = 7.6 (2 dec.)
or
x = 2.354... = 2.4 (2 dec.)

2010-10-10 15:14:44 補充：
2 dec. should be 1 dec.

Q.1:
15^2+x^2=(3x-7)^2(Pyth. Thm.)
x^2+225=9x^2-42x+49
8x^2-42x-176=0
4x^2-21x-88=0
(x-8)(4x+11)=0
x=8 or -2.75(recj.)
so,the length of the shortest stick is 8CM.

Q.2:
a) a=x(20-2x)/2
a=x(10-x)
a=10x-x^2

b) When a=20,
20=10x-x^2
x^2-10x+20=0
x=7.24 or 2.76(corr. to 3 sig. fig.)

When a=18
18=10x-x^2
x^2-10x+18=0
X=7.65 or 2.35(corr. to 3 sig. fig.)

Q1) find the length of the shortest stick
by畢氏定理
x^2+15^2=(3x-7)^2
x^2+225=9x^2-42x+49
8x^2-42x-176=0
(x-8)(4x+11)=0
x=8 or x=-11/4 (rejected)

Q2a) The width of ABCD=a/x
length x width = perimeter
[x+(a/x)]2=20
x+(a/x)=10
x^2+a=10x
a=10x-x^2

b) If a=20
20=10x-x^2
x^2-10x+20=0
x=7.2 or x=2.8

If =18
18=10x-x^2
x^2-10x+18=0
x=7.6 or x=2.4