Question: User: (i) If $x+y=13$ ,$xy=61 4$ find the value of $(x-y)$. solve
1. **State the problem:**
We are given two equations: $x+y=13$ and $xy=61 4$. We need to find the value of $x-y$.
2. **Clarify the product:**
The product $xy=61 4$ seems to have a space. Assuming it means $xy=614$.
3. **Recall the identity:**
We know that $(x-y)^2 = (x+y)^2 - 4xy$.
4. **Substitute the known values:**
$$(x-y)^2 = 13^2 - 4 \times 614$$
5. **Calculate:**
$$(x-y)^2 = 169 - 2456 = -2287$$
6. **Interpretation:**
Since $(x-y)^2$ is negative, there is no real number solution for $x-y$ under these values.
7. **Conclusion:**
The value of $x-y$ is not a real number because the square of $x-y$ is negative.
**Final answer:**
$$x-y = \pm \sqrt{-2287} = \pm i\sqrt{2287}$$