SOLUTION: The sum of the areas of two squares is 225. If one square is 3 units larger per side then the other, find the sizes of the squares.

Question 54370: The sum of the areas of two squares is 225. If one square is 3 units larger per side then the other, find the sizes of the squares.
Found 2 solutions by checkley71, ankor@dixie-net.com:
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
X^2+(X+3)^2=225 OR X^2+X^2+6X+9=225 OR 2X^2+6X+9-225=0 OR 2X^2+6X-216 OR
X^2+3X-108=0 OR (X-9)(X+12)=0 OR X-9=0 OR X=9 & X+12=0 OR X=-12
THUS THE SMALLER SQUARE IS 9 ON A SIDE & THE LARGER IS 9+3=12 AND THE SUM OF THE AREAS ARE 9^2+12^2=225 OR 81+144=225 OR 225=225
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
" one square is 3 units larger per side then the other, find the sizes of the squares."
:
Small square side = x; then larger square side = (x+3)
:
:
"The sum of the areas of two squares is 225".
:
x^2 + (x+3)^2 = 225
:
x^2 + (x^2 + 6x + 9 = 225; FOILed (x+3)^2
:
x^2 + x^2 + 6x + 9 - 225 = 0
:
2x^2 + 6x - 216 = 0
:
x^2 + 3x - 108 = 0; Simplified, divide equation by 2
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Factors to:
(x + 12)(x - 9) = 0
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x = +9, positive solution is what we want here
:
:
Check:
x^2 + (x+3)^2 = 225
9^2 + (9+3)^2 =
81 + 144 = 225
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