1. **Stating the problem:** Solve the system of equations: $$xy = 2 \quad \text{(Equation ①)}$$ $$x^2 + y^2 = 4 \quad \text{(Equation ②)}$$ 2. **Use the given approach:** Multiply Equation ① by 2 and add to Equation ②: $$2xy + (x^2 + y^2) = 4 + 4 = 8$$ Recall that: $$x^2 + 2xy + y^2 = (x + y)^2$$ So, $$ (x + y)^2 = 8 $$ Taking square roots: $$ x + y = \pm 2\sqrt{2} \quad \text{(Equation ③)}$$ 3. **Form quadratic equations:** Treating $x$ and $y$ as roots of a quadratic equation with sum and product: Sum of roots $= x + y = \pm 2\sqrt{2}$ Product of roots $= xy = 2$ The quadratic equation is: $$ t^2 - (x+y)t + xy = 0 $$ Substitute values: For $x + y = 2\sqrt{2}$: $$ t^2 - 2\sqrt{2}t + 2 = 0 $$ For $x + y = -2\sqrt{2}$: $$ t^2 + 2\sqrt{2}t + 2 = 0 $$ 4. **Solve the quadratic equations:** Calculate discriminant for the first: $$ \Delta = (2\sqrt{2})^2 - 4 \times 1 \times 2 = 8 - 8 = 0 $$ So, repeated root: $$ t = \frac{2\sqrt{2}}{2} = \sqrt{2} $$ Similarly for the second: $$ \Delta = (2\sqrt{2})^2 - 4 \times 1 \times 2 = 8 - 8 = 0 $$ Repeated root: $$ t = \frac{-2\sqrt{2}}{2} = -\sqrt{2} $$ 5. **Interpretation:** The solutions for $(x,y)$ are pairs where both are equal to $\sqrt{2}$ or both equal to $-\sqrt{2}$. 6. **Final answer:** $$ \boxed{\{(x,y) = (\sqrt{2}, \sqrt{2}), (-\sqrt{2}, -\sqrt{2})\}} $$ These satisfy both original equations.