Question 1170681


The area of a rectangle is {{{53m^2}}}: 

{{{L*W=53}}}

and if the length {{{L}}} of the rectangle is {{{5m}}} less than {{{twice}}} the width {{{W}}}, we have 

{{{L=2W-5}}}....substitute in area

{{{(2W-5)*W=53}}}

{{{2W^2-5W=53}}}

{{{2W^2-5W-53=0}}}


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

{{{W = (-(-5) +- sqrt( (-5)^2-4*2*(-53) ))/(2*2) }}} 

{{{W = (5 +- sqrt( 25+424 ))/4 }}} 

{{{W = (5 +- sqrt( 449 ))/4 }}} 

{{{W = (5 +- 21.19)/4 }}} 

solutions: need only positive

{{{W = (5 + 21.19)/4 }}} 

{{{W = 6.55}}} 

go to

{{{L=2W-5}}} substitute {{{W}}}

{{{L=2*6.55-5}}}

{{{L=8.1}}}


 the dimensions of the rectangle are approximately {{{8.1}}} by {{{ 6.55}}}