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anonymous

  • 4 years ago

I can't figure out what I'm doing wrong. Solve the equation by completing the square to obtain the exact solution. 3xsquare + 6x + 2 = 0

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  1. anonymous
    • 4 years ago
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    3x^2 + 6x + 2 = 0 You should use quadratic formula.

  2. anonymous
    • 4 years ago
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    The problem I'm having is trying to find the correct square root of the problem.

  3. radar
    • 4 years ago
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    \[x ^{2}+2x+2/3\]\[3x ^{2}+6x+2=0\] dividing thru by 3. \[x ^{2}+2x + 1 +2/3 -1=0\] \[(x+1)^{2}=1/3\]\[x+1=\sqrt{1/3}\] x+1=+/-.577 x=-1.577 x=-.423

  4. anonymous
    • 4 years ago
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    Thanks radar. You proved that I was doing it right, my book has the wrong answer in it.

  5. radar
    • 4 years ago
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    That happens sometimes (but seldom) Have you copied the problem correctly?

  6. radar
    • 4 years ago
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    I'll enter the problem in the Wolfram machine to verify solution.

  7. anonymous
    • 4 years ago
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    Yep. the answer they have in the book is\[-3\pm \sqrt{3}\div3\] I tried doing 10 different ways and I couldn't get that But I got the answer you got every way I tried it

  8. radar
    • 4 years ago
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    The Wolfram graphed the root very close to the values we have obtained. I would discuss it with one of your instructors. Good luck with your studies. I think you are correct.

  9. anonymous
    • 4 years ago
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    Thanks

  10. radar
    • 4 years ago
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    Looking at your book solution, it is similar to what we have:\[x+1 = 1/\sqrt{3}\]\[1/\sqrt{3}\times \sqrt{3}/\sqrt{3}=\sqrt{3}/3\]so in effect we have: \[x+1=\sqrt{3}/3\] or\[x=-1\pm \sqrt{3}/3\] or\[-3\pm \sqrt{3}\over 3\] yes Aurianaa, you have the equivalent of the book answer.

  11. radar
    • 4 years ago
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    It just looks different lol, but is the same.

  12. radar
    • 4 years ago
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    They rationalized the denominator.

  13. anonymous
    • 4 years ago
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    Ok. I thought I was doing it wrong, This problem really upset me lol.

  14. radar
    • 4 years ago
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    No need to talk to your instructor. Or maybe discuss the fact the appearance of the book answer was confusing. I believe you are doing well.

  15. radar
    • 4 years ago
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    The problem was requiring you to use "completing the square" method. and as pratu043 suggested it could of been solved using the quadratic.

  16. anonymous
    • 4 years ago
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    Yea true I did that but I just couldn't figure out how to get the book answer when I knew I was doing it right.

  17. radar
    • 4 years ago
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    I hope all of this has not confused you because it appears that you have comprehension of the procedures.

  18. radar
    • 4 years ago
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    Do you see how the two answers are equivalent?

  19. radar
    • 4 years ago
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    Do you see that multiplying that fraction by\[\sqrt{3}\over \sqrt{3}\] did not change its value only its appearance.

  20. radar
    • 4 years ago
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    Some people don't like radicals in the denominator, they rationalize them as your book did.

  21. anonymous
    • 4 years ago
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    Yes i see how they can go both ways.

  22. radar
    • 4 years ago
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    Good. time for my breakfast. good luck with your studies.

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