algebra
posted by Nabil on .
a wire of 16 inches cut into 2 pieces.
Each piece bent into a square.
Find the length of the two pieces so the sum of the areas of the two squares in 10 square inches.

L = Length of a wire
x = Length of the first pice
y = Length of the cecond pice
x / 4 = Length of the first square
y / 4 = Length of the second square
L = 16 in
16 = x + y
y = 16  x
Area of the first pice = ( x / 4 ) ^ 2
Area of the second pice = ( y / 4 ) ^ 2
( x / 4 ) ^ 2 + ( y / 4 ) ^ 2 = 10
x ^ 2 / 16 + y ^ 2 / 16 = 10
( x ^ 2 + y ^ 2 ) / 16 = 10 Multiply both sides with 16
x ^ 2 + y ^ 2 = 160
x ^ 2 + ( 16  x ) ^ 2 = 160
( Remark: ( 16  x ) ^ 2 = 16 ^ 2  32 x + x ^ 2
x ^ 2 + 16 ^ 2  32 x + x ^ 2 = 160
2 x ^ 2 + 256  32 x = 160
2 x ^ 2 + 256  160  32 x = 0
2 x ^ 2 + 96  32 x = 0
2 x ^ 2  32 x + 96 = 0 Divide both sides with 2
x ^ 2  16 x + 48 = 0
The exact solutions are:
x = 12
and
x = 4
Length of a wire = 16 in
y = 16  x
When: x = 12 ; y = 16  12 = 4
When x = 4 ; y = 16  4 = 12
Length of the two pieces :
12 in
and
4 in
Proof:
( 12 / 4 ) ^ 2 + ( 4 / 4 ) ^ 2 = 10
3 ^ 2 + 1 ^ 2 = 10
9 + 1 = 10 in ^ 2
P.S.
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x ^ 2  16 x + 48 = 0
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x ^ 2  16 x + 48 = 0
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