Question

Jamar wants to generate a Pythagorean triple when x equals 12 and y equals 9 using the polynomial identity (x2+y2)2=(x2−y2)2+(2xy)2
. Which of the following is one of the values of the Pythagorean triple?(1 point)
Responses

225
225

441
441

3,969
3,969

108

Answer 1

To find the Pythagorean triple generated by the polynomial identity \( (x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 \), we can start by calculating the values using \( x = 12 \) and \( y = 9 \).

1. Calculate \( x^2 \) and \( y^2 \):
- \( x^2 = 12^2 = 144 \)
- \( y^2 = 9^2 = 81 \)

2. Calculate \( x^2 + y^2 \):
- \( x^2 + y^2 = 144 + 81 = 225 \)

3. Calculate \( (x^2 + y^2)^2 \):
- \( (x^2 + y^2)^2 = 225^2 = 50625 \)

4. Calculate \( x^2 - y^2 \):
- \( x^2 - y^2 = 144 - 81 = 63 \)

5. Calculate \( (x^2 - y^2)^2 \):
- \( (x^2 - y^2)^2 = 63^2 = 3969 \)

6. Calculate \( 2xy \):
- \( 2xy = 2 \times 12 \times 9 = 216 \)

7. Calculate \( (2xy)^2 \):
- \( (2xy)^2 = 216^2 = 46656 \)

Now verify the polynomial identity:
\[
(x^2 - y^2)^2 + (2xy)^2 = 63^2 + 216^2 = 3969 + 46656 = 50625
\]

Since both sides equal \( 50625 \), we confirm the identity holds true. In a Pythagorean triple form \( a^2 + b^2 = c^2 \), we have:

- \( (x^2 + y^2) = 225 \) (this will be \( c^2 \))
- \( (x^2 - y^2) = 63 \) (this will be \( a \))
- \( 2xy = 216 \) (this will be \( b \))

Now we choose one of the values of the Pythagorean triple from the provided responses. The largest value \( 225 \) is the \( c^2 \) which is part of the Pythagorean triple.

Thus, one of the values of the Pythagorean triple is:

**225**