Question 163715
<pre><font size = 4 color = "indigo"><b>
Plug in 36 for the length L into

{{{L/W = W/(L-W)}}}
{{{36/W = W/(36-W)}}}

Cross-multiply:

{{{36(36-W)=W^2}}}

{{{1296-36W=W^2}}}

Get 0 on the left, by adding -1296
and +36W to both sides:

{{{0=W^2+36W-1296}}}

Swap sides:

{{{W^2+36W-1296=0}}}

That doesn't factor, so we use the
quadratic formula:

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

 with

{{{matrix(1,7, x=W, ",", a=1,",", b=36,",", c=-1296)}}}

{{{W = (-(36) +- sqrt( (36)^2-4*(1)*(-1296) ))/(2*(1)) }}} 

{{{W = (-36 +- sqrt( 6480 ))/2 }}}

{{{W = (-36 +- sqrt(1296*5))/2 }}}

{{{W = (-36 +- sqrt(1296)sqrt(5))/2 }}}

{{{W = (-36 +- 36sqrt(5))/2 }}}

Write as the sum of two fractions:

{{{W = -36/2 +- 36sqrt(5)/2 }}}

{{{W = -18 +- 18sqrt(5) }}}

Simplify the second term by dividing 18 and 2 by 2

{{{W = -18 +- 18sqrt(5) }}}

That's OK like that or you may factor out
9:

{{{W = 18(-1 +- sqrt(5)) }}}

Using the plus:

{{{W = 18(-1 + sqrt(5)) = 22.24922359}}}, approximately.

Using the minus:

{{{W = 9(-1 - sqrt(5)) = -58.24922359}}}, approximately.

We can discard the negative answer, since a rectangle
cannot have a negative side, and the width is given by:

{{{W = 18(-1 + sqrt(5)) = 22.24922359}}} inches, approximately.

Edwin</pre>