Question 921135
 a length of {{{L=(x+3)}}} units and width {{{W=(x-4)}}} if the area of the frame is {{{A=4}}} sq units 

{{{A=L*W}}} 

{{{4=(x+3)*(x-4)}}}

{{{4=x^2-4x+3x-12}}}

{{{4=x^2-x-12}}}

{{{0=x^2-x-12-4}}}

{{{0=x^2-x-16}}}........use quadratic formula

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

{{{x = (-(-1) +- sqrt( (-1)^2-4*1*(-16) ))/(2*1) }}} 

{{{x = (1 +- sqrt( 1+64 ))/2 }}} 


{{{x = (1 +- sqrt(65 ))/2 }}}

{{{x = (1 +- 8.06225774829855)/2 }}}

{{{x = (1 +- 8.06)/2 }}}

solutions:

{{{x = (1 + 8.06)/2 }}}

{{{x = 9.06/2 }}}

{{{x = 4.53}}}

or

{{{x = (1- 8.06)/2 }}}

{{{x =-7.06/2 }}}

{{{x = -3.53}}}..........we don't need this solution because the length and width cannot be negative number

 {{{L=(x+3)}}} units => {{{L=(4.53+3)}}} units=> {{{L=7.53}}} units

and 

{{{W=(x-4)}}} units=>{{{W=(4.53-4)}}} units=>{{{W=0.53}}} units

check the area

{{{4=7.53*0.53}}} 

{{{4=3.9909}}} 

{{{4=4}}}