Question 962358


The length of a rectangular patio is {{{8ft}}} less than twice it's width. 
{{{l=2w-8}}}...eq.1

The area of the patio is {{{A=280ft^2}}}. 

we know that area is {{{A=l*w}}}

so, {{{l*w=280ft^2}}}...now substitute {{{l}}} from eq.1

{{{(2w-8)*w=280}}}.........solve for {{{w}}}

{{{2w^2-8w=280}}}.......both sides divide by {{{2}}}

{{{w^2-4w=140}}}

{{{w^2-4w-140=0}}}.......write {{{-4w}}} as {{{10w-14w}}}

{{{w^2+10w-14w-140=0}}}......group

{{{(w^2+10w)-(14w+140)=0}}}

{{{w(w+10)-14(w+10)=0}}}

{{{(w-14) (w+10) = 0}}}

since we need the width, we can use only positive solution which is

 

now, go back to eq.1 and calculate {{{l}}}

{{{l=2w-8}}}...eq.1
{{{l=2*14-8}}}
{{{l=28-8}}}
{{{highlight(l=20ft)}}}

so, the dimensions of the patio are {{{highlight(l=20ft)}}} and {{{highlight(w=14ft)}}}