Question 887175
A rectangle, with a diagonal of length <i> variable c</i>, 10 inches,has length two inches longer than  it's width, <i> L=w+2</i>.  what is the length of the rectangle? 


Pythagorean Theorem will help.  

{{{highlight_green(w^2+(w+2)^2=10^2)}}}, but you could rework the equation to use L and not w.  But continuing,
{{{w^2+w^2+4w+4=100}}}
{{{2w^2+4w=96}}}
{{{w^2+2w=48}}}
{{{w^2+2w-48=0}}}
{{{(w-6)(w+8)=0}}}
The meaningful answer for w is {{{highlight(w=6)}}}.
This will mean for L,
{{{L=w+2}}}
{{{L=6+2}}}
{{{highlight(L=8)}}}



This part was obviously wrong.....
{{{cross(2w^2+4w+16-100=0)}}}----THIS was the mistake.
{{{2w^2+4w-84=0}}}
{{{w^2+2w-42=0}}}

Discriminant, {{{cross(4-4*(-42)=4+4*42=172)}}}
{{{172=43*4}}}
-
{{{w=(-2+sqrt(43*4))/2}}}---Choosing the PLUS form, because the negative solution of w has no meaning for us here....
{{{w=(-2+2sqrt(43))/2}}}
{{{cross(w=-1+sqrt(43))}}}.


The description gave L=w+2.
{{{L=-1+sqrt(43)+2}}}
{{{cross(L=1+sqrt(43))}}}