Question 1144267
.


            For this problem,  I will show you two ways  (two methods)  of solution.


            First way is  TRADITIONAL.  You may find it everywhere,  and it is boring.



<pre>
If x is the width, then the length is (x+2) and the area equation is


    x*(x+2) = 2  square feet,


    x^2 + 2x - 2 = 0.


Use the quadratic formula


    {{{x[1,2]}}} = {{{(-2 +- sqrt((-2)^2 - 4*1*(-2)))/2}}} = {{{(-2 +- sqrt(12))/2}}} = {{{(-2 +- 2*sqrt(3))/2}}} = {{{-1 +- sqrt(3)}}}.


The width should be positive, so only positive root  x = {{{-1 + sqrt(3)}}}  is the solution for the width.


<U>ANSWER</U>.  The width is  W = {{{sqrt(3)-1}}}.  The length is  L = W+2 = {{{sqrt(3)+1}}}.
</pre>


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The other method is fresh as a matutinal dawn in May, &nbsp;unexpected and elegant. 


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;You may learn it only from me at this forum and in this site -- and practically nowhere else.



<pre>
Let "x" be an unknown value on number line exactly half-way between the length L and the width W values of the rectangle.


Then, OBVIOUSLY,  x = W + 1 = L - 1, and the area is

    L*W = (x+1)*(x-1) = 2,    or


          {{{x^2 - 1}}} = 2,  i.e.

          {{{x^2}}} = 2 + 1 = 3;

     hence,  x = {{{sqrt(3)}}}.  


Thus the dimensions of the rectangle are  W = x-1 = {{{sqrt(3)-1}}}  (the width)  and  L = x+1 = {{{sqrt(3)+1}}}  (the length).


You got the same answer, in a quick and simple manner.
</pre>

See the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/How-to-solve-the-problem-on-quadratic-equation-mentally-and-avoid-boring-calculations.lesson>HOW TO solve the problem on quadratic equation mentally and avoid boring calculations</A> 


&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Surface-area/Problems-on-the-area-and-the-perimeter-of-a-rectangle.lesson>Problems on the area and the dimensions of a rectangle</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Surface-area/Three-methods-to-find-the-dimensions-of-a-rectangle-when-the-givens-are-its-perimeter-and-the-area.lesson>Three methods to find the dimensions of a rectangle when its perimeter and the area are given</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Surface-area/Three-methods-to-find-the-dims-of-a-rectangle-when-its-area-and-the-diff-of-dims-are-given.lesson>Three methods to find the dimensions of a rectangle when its area and the difference of two dimensions are given</A> 

in this site,


where you will find many other similar solved problems &nbsp;(your &nbsp;TEMPLATES)&nbsp; with detailed explanations.


--------------


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Come again soon to this forum to learn more (!)