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Prove that when x > 1, a triangle with side lengths a = x2 − 1, b = 2x, and c = x2 + 1 is a right triangle. Use the Pythagorean theorem and the given side lengths to create an equation. Use the equation to show that this triangle follows the rule describing right triangles. Explain your steps.

40 points!!!

Respuesta :

wegnerkolmp2741o

Answer:

x^2 + 2x^2 + 1 =  x^4 + 2x^2 +1

Step-by-step explanation:

pythagorean theorem

a^2 + b^2 = c^2  where a and b are the legs

substitute what we know

(x^2 -1)^2 + (2x)^2 = (x^2+1)^2

FOIL

(x^2 -1)^2 = (x^2-1 ) (x^2-1) =  x^4 -x^2 - x^2 +1 = x^4 -2x^2 +1

(2x)^2 = 4x^2

(x^2+1)^2 = (x^2+1 ) (x^2+1) =  x^4 +x^2 +x^2 +1= x^4 + 2x^2 +1

substitute these back in

x^4 -2x^2 +1 +4x^2 = x^4 + 2x^2 +1

combine like terms

x^2 + 2x^2 + 1 =  x^4 + 2x^2 +1

Since the left and right side are equal, this is a right triangle


jcherry99

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

Let x = 5

  • (x^2 - 1) = 5^2 - 1
  • 25 - 1 = 24
  • 2x  = 2*5 = 10
  • x^2 + 1 = 5^2 + 1
  • 25 + 1
  • 26

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.

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