Question 1188579

a rectangular field with a length {{{L}}} twice its width {{{W}}}
{{{L=2W}}}......eq.1

If {{{4}}} feet are taken from the length and {{{3}}} feet are taken from the width, the area of the field is {{{150}}} square feet

{{{(L-4)(W-3)=150}}}...........eq.2, substitute {{{L}}}

{{{(2W-4)(W-3)=150}}}

{{{2W^2 - 10W + 12=150}}}

{{{2W^2 - 10W + 12-150=0}}}

{{{2W^2 - 10W -138=0}}}.........divide by {{{2}}}

{{{W^2 - 5W -69=0}}}

{{{W^2 - 5W -69=0}}}

using quadratic formula we get:

{{{W = 5/2 + sqrt(301)/2=11.18 }}}
or  
{{{W = 5/2 - sqrt(301)/2=-6.1746757}}}-> disregard negative solution


{{{L=2*11.18=22.36}}}

 the area of the original field is:

{{{A=L*W}}}

{{{A=22.36*11.18}}}

{{{A=249.99}}}

rounded

{{{A=250}}}