If $$ x^{2}+y^{2}=80 $$ and $$ x y=12 $$, find the value of $$ (x-y)^{2} $$

Question

UpStudy AI · Accepted Answer

To find the value of $$ (x-y)^2 $$, we can use the identity:

$$
(x-y)^2 = x^2 - 2xy + y^2
$$

We already know from the problem statement that:

1. $$ x^2 + y^2 = 80 $$
2. $$ xy = 12 $$

We can substitute these values into the identity. First, we need to calculate $$ -2xy $$:

$$
-2xy = -2 \cdot 12 = -24
$$

Now we can substitute $$ x^2 + y^2 $$ and $$ -2xy $$ into the equation for $$ (x-y)^2 $$:

$$
(x-y)^2 = x^2 + y^2 - 2xy = 80 - 24
$$

Calculating this gives:

$$
(x-y)^2 = 80 - 24 = 56
$$

Thus, the value of $$ (x-y)^2 $$ is

$$
\boxed{56}
$$
The value of $$ (x-y)^2 $$ is 56.

If \( x^{2}+y^{2}=80 \) and \( x y=12 \), find the value of \( (x-y)^{2} \)