Question 132293
Let L=length, W=width



Since the "width of a rectangular is 4ft. less than the length", this means we have the equation:


{{{W=L-4}}}



Also since the "area is 5ft.^2", we have the second equation



{{{A=L*W}}} ----> {{{5=L*W}}} Notice how I plugged in {{{A=5}}}



{{{5=L*(L-4)}}} Now plug in {{{W=L-4}}}



{{{5=L^2-4L}}} Distribute



{{{0=L^2-4L-5}}} Subtract 5 from both sides





{{{0=(L-5)(L+1)}}} Factor the right side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:

{{{L-5=0}}} or  {{{L+1=0}}} 


{{{L=5}}} or  {{{L=-1}}}    Now solve for L in each case



So our answer is 

 {{{L=5}}} or  {{{L=-1}}} 




However, since a negative length doesn't make any sense, the only possible solution is 


{{{L=5}}}



So the length is 5 ft



Now go back to the equation {{{W=L-4}}}



{{{W=5-4}}} Plug in {{{L=5}}}



{{{W=1}}} Subtract



So the width is 1 ft