<PRE><B>Question 456980</B>
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&#39;s true
16^2 = 4(8^2)
256 = 4(64)



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