Evaluate Xy 6D46A9

1. **State the problem:** We need to find $xy$ given different conditions involving $x+y$, $x-y$, $x^2+y^2$, and $(x-y)^2$. 2. **Recall the formula:** $$xy = \frac{(x+y)^2 - (x^2 + y^2)}{2}$$ and also $$(x-y)^2 = x^2 - 2xy + y^2$$ which can be rearranged to find $xy$: $$xy = \frac{x^2 + y^2 - (x-y)^2}{2}$$ --- ### Problem 1: Given $x + y = \sqrt{6}$, $x^2 + y^2 = 4$ 3. Use the formula: $$xy = \frac{(x+y)^2 - (x^2 + y^2)}{2} = \frac{(\sqrt{6})^2 - 4}{2} = \frac{6 - 4}{2} = \frac{2}{2} = 1$$ --- ### Problem 2: Given $x - y = \sqrt{6}$, $x^2 + y^2 = 4$ 4. Use the rearranged formula: $$xy = \frac{x^2 + y^2 - (x-y)^2}{2} = \frac{4 - (\sqrt{6})^2}{2} = \frac{4 - 6}{2} = \frac{-2}{2} = -1$$ --- ### Problem 3: Given $x - y = \sqrt{3}$, $x^2 + y^2 = 5$ 5. Use the rearranged formula: $$xy = \frac{5 - (\sqrt{3})^2}{2} = \frac{5 - 3}{2} = \frac{2}{2} = 1$$ --- ### Problem 4: Given $x + y = \sqrt{10}$, $(x - y)^2 = 2$ 6. First find $x^2 + y^2$ using: $$(x+y)^2 = x^2 + 2xy + y^2$$ So, $$x^2 + y^2 = (x+y)^2 - 2xy = (\sqrt{10})^2 - 2xy = 10 - 2xy$$ 7. Use the rearranged formula for $xy$: $$xy = \frac{x^2 + y^2 - (x-y)^2}{2}$$ Substitute $x^2 + y^2 = 10 - 2xy$ and $(x-y)^2 = 2$: $$xy = \frac{(10 - 2xy) - 2}{2} = \frac{8 - 2xy}{2} = 4 - xy$$ 8. Solve for $xy$: $$xy = 4 - xy$$ $$xy + xy = 4$$ $$2xy = 4$$ $$xy = 2$$ --- **Final answers:** - For $x + y = \sqrt{6}$, $x^2 + y^2 = 4$, $xy = 1$ - For $x - y = \sqrt{6}$, $x^2 + y^2 = 4$, $xy = -1$ - For $x - y = \sqrt{3}$, $x^2 + y^2 = 5$, $xy = 1$ - For $x + y = \sqrt{10}$, $(x - y)^2 = 2$, $xy = 2$