Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

jhonyy9

  • 3 years ago

how is it possible ? x=(pi +3)/2 2x=pi +3 2x(pi -3)=(pi +3)(pi -3) 2pi*x -6x =(pi)^2 -9 9 -6x =(pi)^2 -2pi*x 9 -6x +x^2 =(pi)^2 -2pi*x +x^2 (3 -x)^2 =(pi -x)^2 3-x = pi -x pi =3

  • This Question is Closed
  1. radar
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Everything looks legal, maybe when taking sq roots there is something that gets twisted.

  2. estudier
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If you take the negative root, u get the start equation.....

  3. cwrw238
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    look at the 3rd line if pi = 3 then both sides of equation = 0

  4. AnimalAin
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Didn't consider the plus or minus when taking the square root. We get \[3-x= \pm(\pi-x) \implies 3-x=\pi - x~or~3-x=x-\pi\] The first equation generates the contradiction from the problem. The second generates the value of x given at the get-go.

  5. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy