Question 1050464
Let the smaller side be x, the larger side y.
{{{y=4+x}}}
and
{{{y^2=2x^2-4}}}
Substituting,
{{{(4+x)^2=2x^2-4}}}
{{{16+8x+x^2=2x^2-4}}}
{{{x^2-8x-20=0}}}
{{{(x-10)(x+2)=0}}}
Only one solution since x must be positive,
{{{x-10=0}}}
{{{x=10}}}
So then,
{{{y=4+10=14}}}
Verifying,
{{{14^2=2*10^2-4}}}
{{{196=2*100-4}}}
{{{196=200-4}}}
{{{196=196}}}
True, good solution.