Question 811149


let's the length of a poster be {{{a}}} and the width {{{b}}}

the area is {{{A=a*b}}}.......eq.1

 if the length is {{{10cm}}} longer than its width, then

{{{a=b+10cm}}}.........eq.2

 If the area of the poster is {{{A=1200cm^2}}}, then from eq.1

{{{a*b=1200cm^2}}}......substitute {{{a}}} from eq.2

{{{(b+10cm)*b=1200cm^2}}}.....solve for {{{b}}}to find the width of the poster

{{{b^2+10b*cm=1200cm^2}}}

{{{b^2+10b*cm-1200cm^2=0}}}....use quadratic formula


{{{b = (-10cm +- sqrt((10cm)^2-4*1*(-1200cm^2) ))/(2*1) }}}

{{{b = (-10cm +- sqrt(100cm^2+4800cm^2 ))/2 }}}

{{{b = (-10cm +- sqrt( 4900cm^2 ))/2 }}}

{{{b = (-10cm +- 70cm )/2 }}}....we need only positive solution because the width cannot be negative 

{{{b = (-10cm + 70cm )/2 }}}

{{{b = 60cm /2 }}}

{{{highlight(b = 30cm)}}}....the width of the poster

{{{highlight(a=40cm)}}}........the length of the poster