Question 190793
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That would be <i><b>no</b></i> and <i><b>no</b></i>, refering to your two attempts.


Let <i><b>x</b></i> represent the measure of the side of the smallest square.


Then <i><b>x + 1</b></i> must represent the measure of the side of the middle-sized square, and <i><b>x + 3</b></i> must represent the measure of the side of the largest square.


The area of the smallest square: *[tex \LARGE  x^2]


The area of the middle-sized square: *[tex \LARGE  (x+1)^2] 


The area of the largest square: *[tex \LARGE  (x+3)^2] 


And the sum of these areas is *[tex \LARGE  38\ km^2]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2 + (x + 1)^2 + (x + 3)^2 = 38]


Expand the binomials:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2 + x^2 + 2x + 1 + x^2 + 6x + 9 = 38]


Collect terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  3x^2 + 8x - 28 = 0]


You should be able to take it from here.  This thing is ugly, but it does factor, but you might just want to use the quadratic formula instead of messing around trying to find the factors.  You will get two roots one of which will be negative.  Exclude that root because you are trying to find the measure of something.  The positive root will be the length of the side of the smallest square.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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