Question: User: (i) If $x+y=13$ ,$xy=\frac{25}{4}$ find the value of $(x-y)$. how it it $4xy$ and not $2xy$
1. **State the problem:** We are given $x+y=13$ and $xy=\frac{25}{4}$. We need to find the value of $x-y$.
2. **Recall the formula:** The identity for the square of a difference is:
$$ (x-y)^2 = (x+y)^2 - 4xy $$
This formula is important because it relates $x-y$ to $x+y$ and $xy$.
3. **Calculate $(x-y)^2$ using the given values:**
$$ (x-y)^2 = (13)^2 - 4 \times \frac{25}{4} $$
4. **Simplify the right side:**
$$ (x-y)^2 = 169 - 4 \times \frac{25}{4} = 169 - \cancel{4} \times \frac{25}{\cancel{4}} = 169 - 25 = 144 $$
5. **Find $x-y$ by taking the square root:**
$$ x-y = \pm \sqrt{144} = \pm 12 $$
6. **Answer:** The value of $x-y$ is $\pm 12$.
7. **Explanation about $4xy$ vs $2xy$:**
The formula for $(x-y)^2$ is $(x+y)^2 - 4xy$, not $(x+y)^2 - 2xy$. This is because when you expand $(x-y)^2 = x^2 - 2xy + y^2$, and $(x+y)^2 = x^2 + 2xy + y^2$. Subtracting these two gives:
$$ (x+y)^2 - (x-y)^2 = 4xy $$
So the factor is $4xy$, not $2xy$.