Question 805479
Let {{{ L }}} = the length
Let {{{ W }}} = the width
Let {{{ A }}} = the area
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For a rectangle,  {{{ A = L*W }}}
You are given that {{{ A = 52 }}} ft2, so
(1) {{{ 52 = L*W }}}
You are also given that
(2) {{{ L = 2W - 5 }}}
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Substitute (2) into (1)
(1) {{{ 52 = ( 2W - 5 )*W }}}
(1) {{{ 52 = 2W^2 - 5W }}}
(1) {{{ 2W^2 - 5W - 52 = 0 }}}
Use the quadratic formula
{{{ W = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 2 }}}
{{{ b = -5 }}}
{{{ c = -52 }}}
{{{ W = ( -(-5) +- sqrt( (-5)^2 - 4*2*(-52) )) / (2*2) }}}
{{{ W = ( 5 +- sqrt( 25 + 416 )) / 4 }}}
{{{ W = ( 5 +- sqrt( 441 )) / 4 }}}
{{{ W = ( 5 + 21 ) / 4 }}} ( I can't use the minus square root of {{{ 441 }}} )
{{{ W = 26/4 }}}
{{{ W = 13/2 }}}
and, since
(2) {{{ L = 2W - 5 }}}
(2) {{{ L = 2*(13/2) - 5 }}}
(2) {{{ L = 13 - 5 }}}
(2) {{{ L = 8 }}}
The dimensions are 6.5' x 8'
check:
(1) {{{ 52 = L*W }}}
(1) {{{ 52 = 8*6.5 }}}
(1) {{{ 52 = 52 }}}
OK