Question

Mei wants to generate a Pythagorean triple when x equals 8 and y equals 6 using the polynomial identity (x2+y2)2=(x2−y2)2+(2xy)2
. How does she write the Pythagorean triple, in ascending order?(2 points)
{
,
,
} is a Pythagorean triple.

Answer 1

To generate a Pythagorean triple using the polynomial identity \((x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2\), we first need to calculate the components of the identity using \(x = 8\) and \(y = 6\).

1. Calculate \(x^2\) and \(y^2\):
\[
x^2 = 8^2 = 64
\]
\[
y^2 = 6^2 = 36
\]

2. Calculate \(x^2 + y^2\):
\[
x^2 + y^2 = 64 + 36 = 100
\]

3. Calculate \((x^2 + y^2)^2\):
\[
(x^2 + y^2)^2 = 100^2 = 10000
\]

4. Calculate \(x^2 - y^2\):
\[
x^2 - y^2 = 64 - 36 = 28
\]

5. Calculate \(2xy\):
\[
2xy = 2 \cdot 8 \cdot 6 = 96
\]

Now we can form the Pythagorean triple using the results:
- The first component of the triple:
\[
(x^2 - y^2) = 28
\]

- The second component:
\[
(2xy) = 96
\]

- The hypotenuse:
\[
\sqrt{(x^2 + y^2)} = \sqrt{100} = 10
\]

Putting these together, we have the Pythagorean triple: \(28, 96, 100\).

To write the Pythagorean triple in ascending order:
\[
(10, 28, 96)
\]

However, \(10\) is not part of the original calculation of the triple, and this is not the right approach for forming a Pythagorean triple. The actual values from the identity must be used correctly.

Correctly formed, we actually achieve:
\[
(28, 96, 100)
\]

So, in ascending order:
\[
(28, 96, 100)
\]

Thus, the final answer is:
\[
\{28, 96, 100\}
\]