Question 151709
Okay, since our basis on the {{{highlight(L)ength}}} which is 3 inches longer than the {{{highlight(W)idth}}}, let {{{x=highlight(W)}}}, unknown?
Then, {{{highlight(L)=x+3}}}
Now, adding 2 inches border around it, so the {{{W}}} becomes {{{x+2(top)+2(bottom)}}}. Also the {{{L}}} becomes {{{x+3+2(left)+2(right)}}}
.
Putting this into equation with {{{Area}}} of the photo+border:
{{{A=L*W}}}
{{{108=(x+7)(x+4)}}} ------------> eqn 1
{{{x^2+11x+28-108=0}}} --------> {{{x^2+11x-80=0}}}
By Quadratic Formula: {{{a=1, b=11, c=-80}}}
{{{x=(-b+-sqrt(b^2-4ac))/2a}}}
{{{x=(-11+-sqrt(11^2-4*1*-80))/2*1}}}
{{{x=(-11+-sqrt(121+320))/2}}} -----------> {{{x=(-11+-sqrt(441))/2}}}
{{{x=(-11+-21)/2}}}
{{{x=(-11+21)/2}}} ---------> {{{x=10/2=5}}}
{{{x=(-11-21)/2}}} ----------> {{{x=-32/2=-16}}}
USE {{{x=5inches}}}
For the dimensions of the photo: {{{highlight(W)idth= 5 inches}}}
And, {{{highlight(L)ength=5+3=8inches}}}
In doubt? Go back eqn 1,
{{{108=(5+7)(5+4)}}}
{{{108=(12)(9)}}}
{{{108=108}}}
Thank you,
Jojo