Question 955886
the length of a rectangle {{{L}}} is {{{4}}} more than its width {{{W}}}. 

{{{L=W+4}}}....eq.1

the area of the rectangle {{{A}}} is {{{45}}} square centimeters:
{{{A=45cm^2}}} 

whats is the length and width

{{{A=45cm^2}}}

{{{L*W=45cm^2}}} substitute {{{L=W+4cm}}}

{{{(W+4)*W=45cm^2}}}....solve for {{{W}}}

{{{W^2+4W=45cm^2}}}

{{{W^2+4W-45cm^2=0}}}


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

{{{W = (-4 +- sqrt( 4^2-4*1*(-45cm^2) ))/(2*1) }}}

{{{W = (-4 +- sqrt( 16+180cm^2))/2}}}

{{{W = (-4 +- sqrt( 196cm^2))/2}}}

{{{W = (-4 +- 14cm)/2}}}

solution: we need only positive solution for width

{{{W = (-4 + 14cm)/2}}}

{{{W = 10cm/2}}}

{{{W = 5cm}}}

then the length is

{{{L=5cm+4cm}}}

{{{L=9cm}}}


so, the length is {{{highlight(9cm)}}} and the width is {{{highlight(5cm)}}}