SOLUTION: a 16 inch wire cut into two pieces bent into squares find the length if the sum of the areas is 10 in squared

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Question 258241: a 16 inch wire cut into two pieces bent into squares find the length if the sum of the areas is 10 in squared
Found 2 solutions by Fombitz, ankor@dixie-net.com:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let the two sides be S1 and S2.
1.S1%5E2%2BS2%5E2=10
2.4S1%2B4S2=16
From eq. 2,
S1%2BS2=4
S1=4-S2
Substitute into eq. 1,
S1%5E2%2BS2%5E2=10
%284-S2%29%5E2%2BS2%5E2=10
%2816-8S2%2BS2%5E2%29%2BS2%5E2=10
2S2%5E2-8S2%2B16=10
S2%5E2-4S2%2B3=0
+%28S2-1%29%28S2-3%29=0
Two solutions,
S2-1=0
S2=1
S1=4-S2=3
.
.
.
S2-3=0
S2=3
S1=4-S2=1
One square has a 3 inch side and the other has a 1 inch side.

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = side of one square
Let y = side of the other square
;
Write an equation for each phrase
:
a 16 inch wire cut into two pieces bent into squares
4x + 4y = 16
simplify, divide by 4
x + y = 4
x = (4-y)
:
find the length if the sum of the areas is 10 in squared
x^2 + y^2 = 10
Replace x with (4-y)
(4-y)^2 + y^2 = 10
FOIL
16 - 8y + y^2 + y^2 = 10
Arrange as a quadratic equation
2y^2 - 8y + 16 - 10 = 0
2y^2 - 8y - 6 = 0
Simplify divide by 2
y^2 - 4y + 3 = 0
Factor
(y - 1)(y - 3) = 0
Two solutions
y = 1, then x = 3
y = 3, then x = 1
;
;
Check:
4(1) + 4(3) = 16
3^2 + 1^2 = 10