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)

Answer 1

Use the polynomial identity (x2+y2)2=(x2−y2)2+(2xy)2
to generate a Pythagorean triple when x equals 7 and y equals 4. Write the Pythagorean triple in ascending order.(2 points)

Answer 2

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

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

4. Calculate \( 2xy \):
\[
2xy = 2 \times 8 \times 6 = 96
\]

Now we have:
- \( a = x^2 - y^2 = 28 \)
- \( b = 2xy = 96 \)
- \( c = x^2 + y^2 = 100 \)

Next, the Pythagorean triple will be \( (a, b, c) \):
- With \( a = 28 \), \( b = 96 \), and \( c = 100 \).

To write the Pythagorean triple in ascending order:
\[
(28, 96, 100)
\]

So, the final answer is \( (28, 96, 100) \).

Answer 3

To generate a Pythagorean triple using the polynomial identity \( (x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 \) with \( x = 7 \) and \( y = 4 \), we follow these steps:

1. **Calculate \( x^2 \) and \( y^2 \)**:
\[
x^2 = 7^2 = 49
\]
\[
y^2 = 4^2 = 16
\]

2. **Calculate \( x^2 + y^2 \)**:
\[
x^2 + y^2 = 49 + 16 = 65
\]

3. **Calculate \( x^2 - y^2 \)**:
\[
x^2 - y^2 = 49 - 16 = 33
\]

4. **Calculate \( 2xy \)**:
\[
2xy = 2 \times 7 \times 4 = 56
\]

Now we have:
- The first component \( a = x^2 - y^2 = 33 \)
- The second component \( b = 2xy = 56 \)
- The third component \( c = x^2 + y^2 = 65 \)

The Pythagorean triple is then represented as \( (a, b, c) = (33, 56, 65) \).

To write the Pythagorean triple in ascending order, we find:
\[
(33, 56, 65)
\]

So, the final answer is \( (33, 56, 65) \).