Question 939239
The length {{{L}}} of a rectangular painting is {{{3in}}} longer than its width {{{W}}}. 
{{{L=W+3in}}}

If the diagonal {{{d=15in}}} long, what is the length {{{L}}} of the painting? 

use Pythagorean theorem since the length, the width, and the diagonal of the rectangle form right triangle

{{{d^2=L^2+W^2}}}

{{{(15)^2=(W+3)^2+W^2}}}...solve for {{{W}}}

{{{225=W^2+6W+9+W^2}}}

{{{0=2W^2+6W+9i-225}}}

{{{0=2W^2+6W-216}}}....divide by {{{2}}}

{{{0=W^2+3W-108}}} ...factor, write {{{3W}}} as {{{-9W+12W}}}

{{{W^2-9W+12W-108}}}...group

{{{(W^2-9W)+(12W-108)}}}...factor

{{{W(W-9)+12(W-9)}}}

{{{(W+12)(W-9)=0}}}

solutions:

if {{{(W+12)=0}}}=> {{{W=-12}}}...since we are looking for a width we cannot use negative solution

if {{{(W-9)=0}}}=>{{{highlight(W=9in)}}}...this is our solution

now find the length

{{{L=W+3in}}}

{{{L=9in+3in}}}

{{{highlight(L=12in)}}}

and your answer is 
B.
{{{12}}} inches