Answer:
Step-by-step explanation:
Remark
What an interesting variation on the problem of creating integer values for right triangle solutions. I've never seen it before and yes it does work.
Givens
- a = x^2 - 1
- b = 2x
- c = x^2 + 1
Formula
a^2 + b^2 = c^2
Solution
What you are trying to do is show that the right side and left side will be equal.
(x^2 - 1)^2 + (2x)^2 = (x^2 + 1)^2
x^4 - 2x^2 + 1 + 4x^2 = x^4 + 2x^2 + 1
When you add 4x^2 and - 2x^2 together, you get 2x^2 [on the left side]
x^4 + 2x^2 +1 = x^4+ 2x^2 + 1
Sample Calculation
<em><u>Let x = 5</u></em>
- (x^2 - 1) = 5^2 - 1
- <em>25 - 1 = 24</em>
- <em>2x = 2*5 = 10</em>
- x^2 + 1 = 5^2 + 1
- 25 + 1
- <em>26</em>
24^2 + 10^2 =? 26^2
576 + 100 =? 676
676 = 676 The sample question works.
Conclusion
- The Left hand Side of the equation is equal to the right hand side.
- They represent the integer number solutions to the Pythagorean Theorem, although they don't have to integer values to work. x can be just about any plus value.
- The reason x ≠ 1 is that x^2 - 1 will = 0 and that will never give a value that satisfies the given conditions.