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Use the polynomial identity (x2+y2)2=(x2−y2)2+(2xy)2



to generate a Pythagorean triple when x equals 11 and y equals 4.

Which of the following is one of the values of the Pythagorean triple?

(1 point)
Responses

225
225

137
137

49
49

44

1 answer

To use the polynomial identity \((x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2\) to generate a Pythagorean triple, we'll substitute \(x = 11\) and \(y = 4\).

First, we need to calculate the values of \(x^2\) and \(y^2\):

\[ x^2 = 11^2 = 121 \] \[ y^2 = 4^2 = 16 \]

Now substitute \(x^2\) and \(y^2\) into the equation:

  1. Calculate \(x^2 + y^2\): \[ x^2 + y^2 = 121 + 16 = 137 \]

  2. Calculate \(x^2 - y^2\): \[ x^2 - y^2 = 121 - 16 = 105 \]

  3. Calculate \(2xy\): \[ 2xy = 2 \times 11 \times 4 = 88 \]

Now according to the identity: \[ (x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 \] We have the following values:

  • \(x^2 + y^2 = 137\)
  • \(x^2 - y^2 = 105\)
  • \(2xy = 88\)

Next, let's compute the squares:

  1. \((x^2 + y^2)^2 = 137^2 = 18769\)
  2. \((x^2 - y^2)^2 = 105^2 = 11025\)
  3. \((2xy)^2 = 88^2 = 7744\)

Now checking the Pythagorean nature:

\[ (105)^2 + (88)^2 = 11025 + 7744 = 18769 \]

This confirms the identity holds.

The values \(105\), \(88\), and \(137\) can form a Pythagorean triple.

From your options:

  • 225
  • 137
  • 49
  • 44

The value that is in the Pythagorean triple is 137.