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HelpMe94

  • 2 years ago

Use the Descartes's Rule of Signs to determine the possible numbers of positive and negative zeros of the function: f(x)= -5x^3+X^2-X+5

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  1. mathmate
    • 2 years ago
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    How many times did the expression changed sign?

  2. HelpMe94
    • 2 years ago
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    3

  3. HelpMe94
    • 2 years ago
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    i think

  4. mathmate
    • 2 years ago
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    There is then a MAXIMUM of 3 positive roots, but it could also be 1.

  5. HelpMe94
    • 2 years ago
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    why 1?

  6. mathmate
    • 2 years ago
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    Now flip the sign of the coefficients of the odd powers. How many times does the new expression change sign?

  7. HelpMe94
    • 2 years ago
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    Is that 5X^3+X^2+X+5

  8. mathmate
    • 2 years ago
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    The number of sign changes is the maximum number of positive roots, including complex. If there are two complex roots (as is the case), then there is only one (real) positive root.

  9. HelpMe94
    • 2 years ago
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    ok

  10. HelpMe94
    • 2 years ago
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    so the negative right converted right?

  11. mathmate
    • 2 years ago
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    Yes, the number of sign changes of the new expression (i.e. replace x by -x) gives the MAXIMUM number of negative roots.

  12. HelpMe94
    • 2 years ago
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    So for this one there is none for the negative and 1 for the positive

  13. HelpMe94
    • 2 years ago
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    so how do you know there is one and not 3 for the positive?

  14. mathmate
    • 2 years ago
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    Using the Descartes rule of signs, we do know that there is no negative root. But we do not know if there is ONE or THREE positive roots.

  15. HelpMe94
    • 2 years ago
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    ok so the answer for positive is 1 and 3 and negative is 0

  16. mathmate
    • 2 years ago
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    You can find that out by factoring the expression. A cubic has at least one real root, and we know that it is positive in this case.

  17. mathmate
    • 2 years ago
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    correct.

  18. HelpMe94
    • 2 years ago
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    one other question:

  19. HelpMe94
    • 2 years ago
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    Use the Descartes's Rule of Signs to determine the possible numbers of positive and negative zeros of the function: f(x)=3X^2+2X^2+x+3

  20. HelpMe94
    • 2 years ago
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    how does this work then?

  21. HelpMe94
    • 2 years ago
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    the positive looks like there is none and same with the negative?

  22. mathmate
    • 2 years ago
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    Now, let's see. (1) The expression as is has no change in sign, so what do you conclude? (2) flip the sign of the odd powers to give -3X^3+2X^2-x+3, sign changed 3 times, what do you conclude?

  23. HelpMe94
    • 2 years ago
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    wait why is 3X^2 flipped?

  24. mathmate
    • 2 years ago
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    I supposed you have a typo! It's 3x^3, ... or not?

  25. HelpMe94
    • 2 years ago
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    let me see...

  26. HelpMe94
    • 2 years ago
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    oh srry your right

  27. HelpMe94
    • 2 years ago
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    Use the Descartes's Rule of Signs to determine the possible numbers of positive and negative zeros of the function: f(x)=3X^3+2X^2+x+3

  28. HelpMe94
    • 2 years ago
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    ok so the positive is none then the negative is -3X^3+3X^2-X+3, so 2 and 1

  29. mathmate
    • 2 years ago
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    No positive is correct. Can you repeat the conclusion for negative?

  30. HelpMe94
    • 2 years ago
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    i don't understand

  31. HelpMe94
    • 2 years ago
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    of so negative is 3 and 1

  32. mathmate
    • 2 years ago
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    The number of possible roots always differ by an even number because complex roots, if any, come in pairs. 1 and 2 will not be correct. 1 and 3 would be.

  33. HelpMe94
    • 2 years ago
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    oh ok

  34. mathmate
    • 2 years ago
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    Cool!

  35. HelpMe94
    • 2 years ago
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    cool?

  36. mathmate
    • 2 years ago
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    Perhaps more exactly, 1 OR 3.

  37. HelpMe94
    • 2 years ago
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    ok thanks that makes sense

  38. mathmate
    • 2 years ago
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    You're welcome! :)

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