The identity (x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 can be used to generate Pythagorean triples. What Pythagorean triple could be generated using x = 8 and y = 3?

Question

contestada

Respuesta :

MrRoyal MrRoyal · Answer 1 · 2021-05-28T03:44:19+02:00

Answer:

The triple are: 48, 55 and 73

Step-by-step explanation:

Given

[tex](x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2[/tex]

[tex]x = 8; y=3[/tex]

Required

The Pythagorean triple

We have:

[tex](x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2[/tex]

Substitute: [tex]x = 8; y=3[/tex]

[tex](8^2 + 3^2)^2 = (8^2 - 3^2)^2 + (2*8*3)^2[/tex]

[tex](64 + 9)^2 = (64 - 9)^2 + (48)^2[/tex]

[tex](73)^2 = (55)^2 + (48)^2[/tex]

[tex]5329 = 5329[/tex]

Hence, the triple are: 48, 55 and 73