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theEric

  • 2 years ago

Hi! How do I solve \(x^3-x^2-1=0\) for \(x\). Is this simple, despite Wolfram Alpha's solution, or might it have been written incorrectly? Thanks for any positive input! :)

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  1. blockcolder
    • 2 years ago
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    I don't think the problem is simple.

  2. theEric
    • 2 years ago
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    Thank you! Wolfram Alpha agrees! If the solution isn't found on OpenStudy, I'll try to see what the professor says. The professor is not mine - I was trying to help a friend with this one. Thank you again, @blockcolder :)

  3. satellite73
    • 2 years ago
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    are you supposed to use numerical method like say "newton's method"?

  4. theEric
    • 2 years ago
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    I have no idea! This is in a review to start a semester of calculus II!

  5. satellite73
    • 2 years ago
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    then i guess i have no idea either

  6. Psymon
    • 2 years ago
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    If its a review to start calc 2, then newtons method is a possibility.

  7. theEric
    • 2 years ago
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    Okay, thanks! I know there are methods for third order polynomials, but the polynomials have to be easy to work with in one example I saw. Thank you!

  8. theEric
    • 2 years ago
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    What is Newton's method? Maybe I'll check that out..

  9. theEric
    • 2 years ago
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    Thank you @Psymon , but I'm not sure if they learned that! It is interesting though. I don't understand it yet, it just looks unfamiliar.

  10. Psymon
    • 2 years ago
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    Some calc 1 classes touch on it and others do not. It is an approximation method to find the zeros of a function. The first thing we do is make an initial estimate at where our zero of the function might be close to. This initial guess we can call xn. From there this is the formula: \[x _{n+1}=x _{n}-\frac{ f(x _{n}) }{ f'(x_{n}) }\]Basically we start with xn and use the formula, xn- f(xn)/f'(xn). this result becomes the next xn (hence the x(n+1) in the formula). This calculation is done over and over until the desired degree of accuracy is reached. Usually you can get a good estimate pretty fast, no more than 3, given you picked agood estimate and not one way off.

  11. Psymon
    • 2 years ago
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    For example, a reasonable guess at the zero of this function might be 1.5. If we use that as our initial xn, we can get this: \[1.5-\frac{ f(1.5) }{ f'(1.5) } \approx 1.46667 \]Now I can use this result as my next xn and do the formula again to get: \[1.46667 - \frac{ f(1.46667) }{ f'(1.46667) } \approx 1.46557\] The actual zero is truly at 1.46557. Two uses of the formula and we found the zero./

  12. theEric
    • 2 years ago
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    Thank you very much! I will check with my friend, and I am happy to know this! :) Thank you all!

  13. Psymon
    • 2 years ago
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    Yep, np ^_^

  14. theEric
    • 2 years ago
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    :D

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