Quadratic Equations

1.)
Let the ratio of the two positive integers be x:y=2:3

x/y = 2/3 --- (1)
(x+y)/xy = 1/6 ---(2)

From (1), y=(3x)/2 ---(3) , subs into (2)

{x+[(3x)/2]} / [x(3x)/2] = 1/6

[(2x+3x)/2] * [2/3(x^2)] = 1/6 (約去2)

5x / 3(x^2) = 1/6

30x = 3(x^2)

10x = x^2

(x^2)-10x = 0 (不可直接約去x, 這樣x會少了一個答案)

x(x-10) = 0

x=0 (rejected) or x=10 (x=0要reject係因為題目話呢兩個數係positive integers, 0唔係positive/negative integers)

x=10 , subs into (3)
y=3(10)/2
y=15
The two positive integers are 10 and 15



2.)
Let the tens digit be x and the units digit be y

x=y+2 ---(1)
xy+40 = 10x+y ---(2) (因為題目話xy係一個兩位數既數目, 將呢個數目拆開成兩個數字乘埋, 會比呢個兩位數少40, 所以x*y+40, 會等於個兩位數; 而x本身係十位數, 所以要*10。假設呢個兩位數係96, 拆開黎睇, x=9, y=6, 但其實呢兩位數係等於90+6, 姐係10x+y)

Subs (1) into (2)
[(y+2)y]+40 = 10(y+2)+y
(y^2)+2y+40 = 10y+20+y
(y^2)+2y+40 = 11y+20
(y^2)-9y+20 = 0
(y-5) (y-4) = 0
y = 5 or y=4 , subs into (1)
x=5+2 or x=4+2
x=7 or x=6
Therefore, when x=7 , y=5 , and the 2-digit number is 75 ;
when x=6 , y=4 , and the 2-digit number is 64.

咁樣明唔明=0=?

俾住1,2題你先, 有時間先做埋第3,4題。

2014-10-09 22:46:00 補充：
3.)
Let the smallest circle be circle A, the middle circle be circle B, and the largest circle be circle C.
area of circle A = (5^2)
area of circle B = (5+x)^2
area of circle C = (5+2x)^2

area of white region = area of circle B - area of circle A = (5+x)^2*π - (5^2)π

2014-10-09 22:50:57 補充：
area of shaded region = area of circle A + area of circle C (only shaded region)
=area of circle A + (area of circle C - area of circle B)
= (5^2)π + (5+2x)^2*π - (5+x)^2*π

The area of shaded region is twice of the area of white region

(5^2)π + (5+2x)^2π - (5+x)^2π = 2[(5+x)^2π - (5^2)π]

2014-10-09 22:57:15 補充：
25π + (5+2x)^2π - (5+x)^2π = 2(5+x)^2π -50π

π [(5+2x)^2 - (5+x)^2] = π [2(5+x)^2 - 75]

[25+20x+4(x^2)] - [25+10x+(x^2)] +75 = 2[25+10x+(x^2)]

25+20x+4(x^2) - 25-10x-(x^2) + 75 = 50+20x+2(x^2)

10x+3(x^2)+75 = 50+20x+2(x^2)

x-10x+25 = 0

(x-5)^2 = 0

x=5 (repeated)

2014-10-09 23:03:42 補充：
5+2x=15cm

2014-10-09 23:09:18 補充：
回答果到做左1-3題俾你, 但第4題做唔到, 上面果part畫到, 但之後folding along AP and QC果到, where is point P and point Q?

加題目不知加到何時。

1.
Let x be the largest positive integer,
y be another positive integer
y:x=2:3
y=2x/3

(x+y):(xy)=1:6
6[x+ (2x/3)]=x(2x/3)
10x=2x^2 /3
2x^2 - 30x =0
2x(x-15)=0
x=15 or 0 (rejected, because if x=0, then y<0)

y=2(15)/3=10
∴The two integers are 10,15.

2014-10-02 15:57:16 補充：
2.
Let x,y be the tens digit and the units digit respectively.
x-y=2
y=x-2

xy+40=10x+y
x(x-2)+40=10x+x-2
x^2 - 2x + 40 - 11x + 2=0
x^2 - 13x + 42=0
(x -6)(x+7)=0
x=6 or 7

If x=6,y=6-2=4
If x=7,y=7-2=5
∴64,75

2014-10-02 16:45:00 補充：
我諗可能係類似以下嘅圖片
http://postimg.org/image/5ca40q9pz/

因為咁樣個radius可以計到1個正整數出黎

2014-10-03 19:26:46 補充：
0.0

2.
x=the tens digit
y=the units digit

x-y=2係解做The tens digit of a 2-digit number is greater than the units digit by 2

10x+y係指 the 2-digit number
xy係指 the tens digit 乘 the units digit
(10x+y)-(xy)=40係指the product of the digits is less than the 2-digit number by 40

2014-10-07 15:20:23 補充：
我諗第4題應該要跟住個圖黎計 0.0