SOLUTION: when each side lenth of square is increased 10 inches the resulting areas is four times as greater as it would of been if each side length had been increased by 2 inches determen t

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Question 456980: when each side lenth of square is increased 10 inches the resulting areas is four times as greater as it would of been if each side length had been increased by 2 inches determen the origanl side leght of the squar

Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
Each side length of square is increased 10 inches, the resulting area is four times greater as it would of been, if each side length had been increased by 2 inches.
Determine the original side length of the square
:
Let s = length of original square
:
(s+10)^2 = 4(s+2)^2
FOIL
s^2 + 10s + 10s + 100 = 4(s^2 + 2s + 2s + 4)
s^2 + 20s + 100 = 4(s^2 + 4s + 4)
s^2 + 20s + 100 = 4s^2 + 16s + 16
:
Combine like terms on the left,
s^2 - 4s^2 + 20s - 16s + 100 - 16 = 0
-3s^2 + 4s + 84 = 0
:
multiply by -1, we want the coefficient of s^2 to be positive
3s^2 - 4s - 84 = 0
;
Factors to:
(3s + 14)(s - 6) = 0
the positive solution is what we want here
s = 6 inches, length of the original square
:
:
See if that's true
16^2 = 4(8^2)
256 = 4(64)



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