SOLUTION: A piece of wire L inches long is cut into two pieces. Each piece is then bent to form a square. If the sum of the two areas is 5L squared/128 (sorry, I don't know how to make the

Question 79735: A piece of wire L inches long is cut into two pieces. Each piece is then bent to form a square. If the sum of the two areas is 5L squared/128 (sorry, I don't know how to make the squared sign for my L term), how long are the two pieces of wire? Answer in terms of L.
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Area(A) of a square=the length of a side squared or A=S^2
Let x=length of one piece of wire
The length of each side of this square would be x/4 inches;
The area would be x^2/16 sq in
Then L-x= length of the other piece
The length of each side of this square would be (L-x)/4 inches;
The area would be (L-x)^2/16 sq inches
Now we are told that the sum of these areas equals 5L^2/128. So our equation to solve is:
x^2/16+(L-x)^2/16=5L^2/128 multiply each term by 128 to get rid of fractions
128x^2/16+128(L-x)^2/16=128(5L^2)/128 simplify
8x^2+8(L^2-2Lx+x^2)=5L^2 get rid of parens
8x^2+8L^2-16Lx+8x^2=5L^2 subtract 5L^2 from both sides and collect like terms
16x^2-16Lx+3L^2=0 quadratic in standard form
A=16
B=-16L
C=+3L^2
Solve using the quadratic formula:

and

What this tells us is that one piece is L/4 and the other is 3L/4 and visa versa.
CHECK
Area of one square is (1/4 of 3L/4)^2=(3L/16)^2=9L^2/256
Area of other square is (1/4 of L/4)^2=(L/16)^2 =L^2/256
9L^2/256+L2/256=10L^2/256 which equals 5L^2/128 and this checks

Hope this helps---ptaylor

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