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baldymcgee6

  • 2 years ago

Prove that this is true: x*sqrt(x+1) - sqrt(x^2+1)*(x-1) = x^2 - 2x -1

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  1. baldymcgee6
    • 2 years ago
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    \[x*\sqrt(x+1) - \sqrt(x^2+1)*(x-1) = x^2 - 2x -1\]

  2. Jemurray3
    • 2 years ago
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    It isn't.

  3. baldymcgee6
    • 2 years ago
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    \[x*\sqrt{x+1} - \sqrt{x^2+1}*(x-1) = x^2 - 2x -1\]

  4. baldymcgee6
    • 2 years ago
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    that's what i thought too...

  5. Jemurray3
    • 2 years ago
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    if you plug in x = 1 I get sqrt(2) = -2.

  6. baldymcgee6
    • 2 years ago
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    k thanks.

  7. calculusfunctions
    • 2 years ago
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    By proving that L.S. = R.S. we're proving that it is an identity. Which implies that the equation is valid for all x except x > -1. Therefore Jemurray is correct.

  8. calculusfunctions
    • 2 years ago
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    Let me ask you though. Do you mean for the (x − 1) to be inside the square root in the second term of the left side of the equation?

  9. Jemurray3
    • 2 years ago
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    Even if he did, it doesn't matter. The fact that the first term on the left is only valid on a restricted domain and the others aren't is enough to show that it can't be an equality.

  10. calculusfunctions
    • 2 years ago
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    I know that but I'm grasping for straws trying to understand what he's talking about or even if asked the question correctly.

  11. baldymcgee6
    • 2 years ago
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    @calculusfunctions, no it is not under the root.

  12. calculusfunctions
    • 2 years ago
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    Are you sure you copied the question correctly?

  13. calculusfunctions
    • 2 years ago
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    Could you please double check because I'd really like to help you if I could.

  14. baldymcgee6
    • 2 years ago
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    yep, don't worry about it. It is very possible that they are not equal.

  15. baldymcgee6
    • 2 years ago
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    In fact it is so possible that it is true that they are not equal. :)

  16. baldymcgee6
    • 2 years ago
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    but thanks for your eagerness to help!

  17. calculusfunctions
    • 2 years ago
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    You're right! If LS ≠ RS for some values of x, then the equation is not an identity.

  18. calculusfunctions
    • 2 years ago
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    No worries. Welcome!

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