有一條quadratic equations 唔係好識

Let y be the length of the longer wire;
so (80-y) is the length of the shorter wire.


From the question, "Each part is bent to form a square ".
As you know, a square has 4 edges.
So 1 edge of those square are y/4 and (80-y)/4 respectively,
and the area of those square are ( y/4 )^2 and [ (80-y)/4 ]^2 respectively.


Hence,
( y/4 )^2 + [ (80-y)/4 ]^2 = 272
y^2/16 + 400 - 10y + y^2/16 = 272
y^2 - 80y + 1024 = 0
( y - 16 )( y - 64 ) = 0
y = 64 or y =16 (rejected)


Therefore the length of the longer wire is 64 cm.

Let 4x cm be the longer wire :

then the shorter wire = 80 - 4x cm

So 4x > 80 - 4x
x > 10


Big square = x²

Small square = (20 - x)²


x² + (20 - x)² = 272

x² + x² - 40x + 400 = 272

2x² - 40x + 128 = 0

x² - 20x + 64 = 0

(x - 4) (x - 16) = 0

2011-09-16 00:09:16 補充：
x = 4 (rejected since x > 10) or x = 16

The longer wire = 4x = 4 * 16 = 64 cm

method:let一個unknown

Let the length of the longer wire be x cm.
Then the total area of two squares=272
(x/4)^2+[(80-x)/4]^2=272
(x^2/16)+[(80-x)^2/16]=272
x^2+(80-x)^2=(272)(16)
x^2+80^2-(2)(80)(x)+x^2=4352
2x^2-160x+6400-4352=0
2x^2-160x+2048=0
x^2-80x+1024=0
(x-16)(x-64)=0
Therefore x=16 or x=64
The longer wire is 64cm.
P.S. The shorter one is exactly 16cm.