Question 1204421
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A question like this has been asked before.


Review this link to see an example how to solve.
<a href="https://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.1197474.html">https://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.1197474.html</a>


If you are still stuck, then read on.


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Draw out a diagram as indicated in the link above.


There will be a large rectangle that is 27 feet by 35 feet. 
This represents the dimensions of the room.


Inside that larger rectangle will be a smaller rectangle with dimensions (27-2x) feet by (35-2x) feet.
x represents the width of the uniform strip. 
This is the buffer gap between the edge of the rug to the edge of the wall.


The area of the smaller rectangle is (27-2x)(35-2x).
Set this equal to 609 because this is the amount of carpet she can buy.


(27-2x)(35-2x) = 609
27(35-2x)-2x(35-2x) = 609
945-54x-70x+4x^2 = 609
945-54x-70x+4x^2 - 609 = 0
4x^2 - 124x + 336 = 0
4(x^2 - 31x + 84) = 0
x^2 - 31x + 84 = 0


We could factor, but I think the quadratic formula is the most efficient pathway. 
This is because we don't have to worry about guess-and-check.


Plug a = 1, b = -31, c = 84 into the quadratic formula below.
{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-(-31)+-sqrt((-31)^2-4(1)(84)))/(2(1))}}}


{{{x = (31+-sqrt(961 - 336))/(2)}}}


{{{x = (31+-sqrt(625))/(2)}}}


{{{x = (31+-  25)/(2)}}}


{{{x = (31+25)/(2)}}} or {{{x = (31-25)/(2)}}}


{{{x = (56)/(2)}}} or  {{{x = (6)/(2)}}}


{{{x = 28}}} or  {{{x = 3}}}


If x = 28, then 27-2x = 27-2*28 = -29, which is not valid. 
The side length must be positive.


If x = 3, then, 
27-2x = 27-2*3 = 21
and
35-2x = 35-2*3 = 29
Both results are positive. 
Therefore x = 3 is valid and practical.
A gap of 3 feet produces a rug that is 21 feet by 29 feet.


As a check, 21*29 = 609, and we have confirmed the answer is correct.


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<font size=4>Answer:</font> The rug should be 21 feet by 29 feet.
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