Question 890126
Original length and width, L and w.


L=w+36.
{{{wL=w(w+36)=5152}}}.
{{{w^2+36w-5152=0}}}
5152=4*4*2*161.
Discrim, 36^2-4*(-5152)=36+4*5152=21904, which is {{{148^2}}}.
{{{w=(-36+148)/2}}}---using general solution of quadratic equation
{{{highlight(w=56)}}}
{{{highlight(L=92)}}}



Double the area.
Let the unknown distances increase by factor x .
{{{(w*x*Lx)=2*5152}}}, what is x ?
{{{wL*x^2=2*5152}}}, we already know wL=5152.
x^2=2
{{{highlight_green(x=sqrt(2))}}}----this is not doubled.


<i>...would its length and width also double?</i>   No.