Examine this set of pythagorean triples from part c. look for a pattern that is true for each triple regarding the difference between the three values that make up the triple. describe this pattern. then see if you can think of another pythagorean triple that doesn't follow the pattern you just described and that can't be generated using the identity (x^2-1) (2x)^2 (x^2+1)^2. explain your findings

did you get the answer for this?

You can check that the numbers on your table are Pythagorean triplets by substituting them in the Pythagorean equation:

6² + 8² = 36 + 64 = 100 = 10²

8² + 15² = 64 + 225 = 289 = 17²

10² + 24² = 100 + 576 = 676 = 26²

12² + 35² = 144 + 1225 = 1369 = 37²

Now, lets look for the pattern:

x-value     Pythagorean

                 triple    

3               (6, 8, 10)            

 6/2 = 3

3² - 1 = 9 - 1 = 8

3² + 1 = 9 + 1 = 10

4               (8, 15, 17)            

 8/2 = 4

4² - 1 = 16 - 1 = 15

 4² + 1 = 16 + 1 = 17

5               (10, 24, 26)            

 10/2 = 5

 5² - 1 = 25 - 1 = 24

24² + 1   = 25 + 1 = 26

6               (12, 35, 37)            

 12/2 = 6

6² - 1 = 36 - 1 = 35

6² + 1   = 36 + 1 = 37

From which you find the pattern: the first number is 2x, the second number is x² - 1, and the third number is x² + 1

(2x)² + (x² - 1)² = (x² + 1)², or

(x² - 1)² + (2x)² = (x² + 1)².

Other example of a Pythagorean triple is (3, 4, 5). x = 2, you end with:

x = 2

2x =  2(2) = 4

x² - 1 = 2² - 1 = 3

x² + 1 = 2² + 1 = 5

(3, 4, 5) also follows the pattern.

Only  right triangles with non-integer sides do not form Pythagorean triples.

Of course you may proof that (x² - 1)² + (2x)² = (x² + 1)² is an identity

Left hand side: (x⁴ - 2x² + 1) + 4x² = x⁴ + 2x² + 1

Right hand side: x⁴ + 2x² + 1

The equation is always true.

At the end, the pattern is true for any Pythagorean triplet

there is no points to select.

step-by-step explanation: