Question 689435

given:

The width {{{W}}} of a rectangle is {{{6m}}} {{{less}}} than its length {{{L}}}. 

{{{W=L-6m}}}

The area {{{A}}} is {{{315m^2}}}.

to find: the length {{{L}}} 


as you know {{{A=L*W}}}.......plug in given values


{{{315m^2=L*(L-6m)}}}

{{{315m^2=L^2-L6m)}}}

{{{L^2-L6m-315m^2=0)}}}.....use quadratic formula to solve for {{{L}}} 


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

{{{L = (-(-6m) +- sqrt( (-6m)^2-4*1*(-315m^2) ))/(2*1) }}}

{{{L = (6m +- sqrt( 36m^2+1260m^2 ))/2 }}}

{{{L = (6m +- sqrt( 1296m^2 ))/2 }}}


{{{L = (6m +- 36m )/2 }}}.........find only positive root because the length cannot be negative


{{{L = (6m + 36m )/2 }}}

{{{L = 42m /2 }}}

{{{highlight(L = 21m ) }}}.....your answer

we can find the width too:

{{{W=L-6m}}}

{{{W=21m-6m}}}

{{{highlight(W=15m)}}}