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vanndaleyrang

Can someone solve this quadratic equation x^2/3-x-5/3?

  • one year ago
  • one year ago

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  1. satellite73
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    is it \[\frac{x^2}{3}-x+\frac{5}{3}=0\]?

    • one year ago
  2. timo86m
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    holy crap use perenthesis

    • one year ago
  3. vanndaleyrang
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    @satellite73 You got it right just change the +5/3 to -5/3

    • one year ago
  4. satellite73
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    if so, start by multiplying everything by \(3\) to get \[x^2-3x-5=0\]

    • one year ago
  5. timo86m
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    \[1/3\,{x}^{2}-x-5/3\] that is what maple translates it to

    • one year ago
  6. satellite73
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    unfortunately i don't think this one factors, as a matter of fact i am sure it does not

    • one year ago
  7. satellite73
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    so you have to use the quadratic formula

    • one year ago
  8. timo86m
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    so complete the square :) Also i think it is a good idea to say expression=0 when you say solve. Other wise we imply you mean = to 0

    • one year ago
  9. vanndaleyrang
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    @timo86m Yeah this is my first time using open study so im trying to my best, lol bare with me. Okay can you show me step by step how to complete the square?

    • one year ago
  10. satellite73
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    use \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] with \(a=1,b=-3,c=-5\) i would not complete the square for this one

    • one year ago
  11. timo86m
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    satallite already started in quadratic formula instead

    • one year ago
  12. vanndaleyrang
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    Okay thats fine

    • one year ago
  13. satellite73
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    if you want to complete the square, which the quadratic formula does for you in one step, you have to start with \[x^2-3x=5\] then take half of 3, which is \(\frac{3}{2}\) and write \[(x-\frac{3}{2})^2=5+\frac{9}{4}=\frac{27}{4}\] and so \[x-\frac{3}{2}=\pm\frac{\sqrt{27}}{2}\] making \[x=\frac{3\pm\sqrt{27}}{2}\]

    • one year ago
  14. timo86m
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    altho i am pretty sure you will have to learn completing the square sooner or later :)

    • one year ago
  15. satellite73
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    except i made a mistake, \(5+\frac{9}{4}=\frac{29}{4}\)

    • one year ago
  16. vanndaleyrang
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    Ok, thank you guys! :) @timo86m @satellite73

    • one year ago
  17. satellite73
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    in any case if the middle term is odd,(-3\) is odd, then it is easier to use the quadratic formula

    • one year ago
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