Question 658834
From {{{x - y = 4}}} we know {{{y = x-4}}}
Substituting into the first equation gives
{{{x^2 + (x-4)^2 = 48}}}
Expanding:
{{{x^2 + x^2 - 8x + 16 = 48}}}
Combining like terms:
{{{2x^2 - 8x + 16 = 48}}}
Subtracting 48 from both sides:
{{{2x^2 - 8x - 32 = 0}}}
Dividing both sides by 2:
{{{x^2 - 4x - 16 = 0}}}
This can be solved using the quadratic equation {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} with a=1, b=-4, and c=-16
{{{x = (-(-4) +- sqrt( (-4)^2-4*1*(-16) ))/(2*1) }}}
{{{x = (4 +- sqrt( 16+64 ))/2 }}}
{{{x = (4 +- sqrt( 80 ))/2 }}}
{{{x = (4 +- sqrt( 16*5 ))/2 }}}
{{{x = (4 +- 4*sqrt( 5 ))/2 }}}
{{{x = 2 +- 2*sqrt( 5 ) }}}
So x can have the values {{{x=2+2*sqrt(5)}}} or {{{x=2-2*sqrt(5)}}}
For {{{x = 2+2*sqrt(5)}}} we have {{{y = x-4 = -2+2*sqrt(5)-2}}}
{{{x*y = (2+2*sqrt(5))(2*sqrt(5)-2) = 4*sqrt(5) - 4 + 20 - 4*sqrt(5) = 16}}}
Checking:
{{{x^2 + y^2 = (2 + 2*sqrt(5))^2 + (-2 + 2*sqrt(5)-2)^2 = (4 + 8*sqrt(5)+20) + (4 - 8*sqrt(5) + 20) = 4+20+20+4 + 8*sqrt(5) - 8*sqrt(5) = 48}}}
And
{{{x*y = (2-2*sqrt(5))(-2-2*sqrt(5)) = -4 - 4*sqrt(5) + 4*sqrt(5) + 20 = 16}}}

For {{{x = 2-2*sqrt(5)}}} we have {{{y = x-4 = -2-2*sqrt(5)}}}
Checking:
{{{x^2 + y^2 = (2 - 2*sqrt(5))^2 + (-2 - 2*sqrt(5)-2)^2 = (4 - 8*sqrt(5)+20) + (4 + 8*sqrt(5) + 20) = 4+20+4+20 - 8*sqrt(5) + 8*sqrt(5) = 48}}}
And
{{{x*y = (2-2*sqrt(5))(-2-2*sqrt(5)) = -4 - 4*sqrt(5) + 4*sqrt(5) + 20 = 16}}}
So for either value of x, x*y = 16