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vanndaleyrang

  • 2 years ago

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

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

  2. timo86m
    • 2 years ago
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    holy crap use perenthesis

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

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

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

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

  7. satellite73
    • 2 years ago
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    so you have to use the quadratic formula

  8. timo86m
    • 2 years ago
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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

  9. vanndaleyrang
    • 2 years ago
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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?

  10. satellite73
    • 2 years ago
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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

  11. timo86m
    • 2 years ago
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    satallite already started in quadratic formula instead

  12. vanndaleyrang
    • 2 years ago
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    Okay thats fine

  13. satellite73
    • 2 years ago
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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 \[x=\frac{3\pm\sqrt{27}}{2}\]

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

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

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

  17. satellite73
    • 2 years ago
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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

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