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use the polynomial identity (x^2+x^y)^2= (x^2-y^2)^2+(2xy)^2 to generate a pythagorean triple when x equals 7 and y equals 4 write the Pythagorean triple in ascending order

3 answers

To use the polynomial identity \((x^2 + x^y)^2 = (x^2 - y^2)^2 + (2xy)^2\) with \(x = 7\) and \(y = 4\), we first need to compute the necessary parts of the identity.

  1. Calculate \(x^2\): \[ x^2 = 7^2 = 49 \]

  2. Calculate \(x^y\) (which is \(x^4\)): \[ x^y = 7^4 = 2401 \]

  3. Now we can calculate the left-hand side of the identity: \[ (x^2 + x^y)^2 = (49 + 2401)^2 = 2450^2 \]

  4. Next, we calculate the right-hand side:

    • First, calculate \(x^2 - y^2\): \[ x^2 - y^2 = 49 - 16 = 33 \]
    • Now, compute \((x^2 - y^2)^2\): \[ (x^2 - y^2)^2 = 33^2 = 1089 \]
    • Now calculate \(2xy\): \[ 2xy = 2 \cdot 7 \cdot 4 = 56 \]
    • Now compute \((2xy)^2\): \[ (2xy)^2 = 56^2 = 3136 \]
    • Finally, add the results from the right-hand side: \[ (x^2 - y^2)^2 + (2xy)^2 = 1089 + 3136 = 4225 \]

Now we can summarize the results:

  • Left-hand side: \(2450^2\) (though we won't directly need this value)
  • Right-hand side: \(4225 = 65^2\)

Now, we have \(a\), \(b\), and \(c\) from the identity:

  • \(a = |x^2 - y^2| = 33\)
  • \(b = 2xy = 56\)
  • \(c = |x^2 + y^2| = 65\)

Therefore, the Pythagorean triple generated is \( (33, 56, 65) \).

Finally, to write the Pythagorean triple in ascending order: \[ (33, 56, 65) \]

Are you sure

Let's go through the problem step by step once more to ensure everything is clear and correct.

Given the polynomial identity: \[ (x^2 + x^y)^2 = (x^2 - y^2)^2 + (2xy)^2 \]

We will substitute \(x = 7\) and \(y = 4\).

Step 1: Calculate \(x^2\) and \(x^y\)

  • Calculate \(x^2\): \[ x^2 = 7^2 = 49 \]

  • Calculate \(x^y = x^4\): \[ x^y = 7^4 = 2401 \]

Step 2: Calculate the left-hand side

Now, compute the left-hand side: \[ (x^2 + x^y)^2 = (49 + 2401)^2 = (2450)^2 \]

Step 3: Calculate the right-hand side

Now compute each part on the right-hand side:

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

  2. Compute \((x^2 - y^2)^2\): \[ (33)^2 = 1089 \]

  3. Calculate \(2xy\): \[ 2xy = 2 \cdot 7 \cdot 4 = 56 \]

  4. Compute \((2xy)^2\): \[ (56)^2 = 3136 \]

  5. Add the two squares together: \[ (x^2 - y^2)^2 + (2xy)^2 = 1089 + 3136 = 4225 \]

Step 4: Verify Pythagorean triple

Now, according to the identity: \[ (x^2 + x^y)^2 = (x^2 - y^2)^2 + (2xy)^2 \] the right-hand side \( (x^2 - y^2)^2 + (2xy)^2 = 4225 = 65^2 \).

Based on the values we calculated:

  • \(a = |x^2 - y^2| = 33\)
  • \(b = 2xy = 56\)
  • \(c = \sqrt{(x^2 + y^2)} = 65\)

Resulting Pythagorean Triple

Therefore, the Pythagorean triple is: \[ (33, 56, 65) \]

Given that you want it in ascending order, it is \( (33, 56, 65) \). The calculations are indeed correct, and this is a valid Pythagorean triple. Thank you for your patience!