Question

Ashley wants to generate a Pythagorean triple when x = 18 and y = 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 324 324 162 162 81 81 729

Answer 1

To generate a Pythagorean triple using the polynomial identity \((x^2+y^2)^2 = (x^2-y^2)^2 + (2xy)^2\), we need to calculate the required terms using \(x = 18\) and \(y = 9\).

First, we calculate \(x^2\) and \(y^2\):

1. \(x^2 = 18^2 = 324\)
2. \(y^2 = 9^2 = 81\)

Now we can calculate the expressions:

1. \(x^2 + y^2 = 324 + 81 = 405\)
2. \((x^2 + y^2)^2 = 405^2 = 164025\)

Next, we calculate \(x^2 - y^2\) and \(2xy\):

1. \(x^2 - y^2 = 324 - 81 = 243\)
2. \(2xy = 2 \cdot 18 \cdot 9 = 324\)

Now we can find \((x^2 - y^2)^2 + (2xy)^2\):

1. \((x^2 - y^2)^2 = 243^2 = 59049\)
2. \((2xy)^2 = 324^2 = 104976\)

Adding these two results together:

\[
59049 + 104976 = 164025
\]

Thus, both sides of the equation match, confirming that a Pythagorean triple can be generated.

The values of the generated Pythagorean triple are:

1. The first side \(= x^2 + y^2 = 405\)
2. The second side \(= |x^2 - y^2| = 243\)
3. The hypotenuse \(= 2xy = 324\)

From the provided options (324, 162, 81, 729), the value that appears in our Pythagorean triple is:

**324**.