How can we show that there are no (x,y) pairs of real numbers that satisfy xy = 3 and (x + y)^2 = 10 ? Intuitively it's obvious but I'm not sure exactly how to show it..
Brute force:
Sub y = 3/x into (x + y)^2 = 10.
After expanding and simplifying you get $\displaystyle x^2 + \frac{9}{x^2} = 4$. Multiply through by x^2 and re-arrange:
$\displaystyle x^4 - 4x^2 + 9 = 0$
$\displaystyle \Rightarrow (x^2 - 2)^2 - 4 + 9 = 0$
$\displaystyle \Rightarrow (x^2 - 2)^2 = -5$.
Therefore no real solutions.