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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)

3 answers

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 \( (28, 96, 100) \).

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)

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 \( (33, 56, 65) \).