Question 180246
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Let the side of the square be <i>s</i>.  Then the area is <i>s</i>² and the perimeter is 4<i>s</i>.  But we know the area to be 5 larger than the perimeter, so:


*[tex \Large s^2 - 5 = 4s]


In standard form:


*[tex \Large s^2 - 4s - 5 = 0]


Factor:


*[tex \Large (s - 5)(s + 1) = 0]


So:


*[tex \Large s - 5 = 0]  &#8594; *[tex \Large s = 5]


or 


*[tex \Large s + 1 = 0]  &#8594; *[tex \Large s = -1]


Exclude the second root because -1 for the length of the side of a square is absurd.  It is an extraneous root caused by squaring the variable while solving the problem.  That leaves *[tex \Large s = 5]


Check:


The area of a square with side length 5 is 25.  The perimeter of a square with side length 5 is 20.  25 is 5 larger than 20.  Checks.



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