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neverforgetvivistee

  • 3 years ago

Can you simplify this further? *Using the quadratic formula to factor x^2-6x-4* The answer is supposed to be 6(+-)√53 / 2 ,but can't you simplify by 2? 6 and 2 go into 2. Then could it be 3(+-)√68/1 or no?

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  1. Hero
    • 3 years ago
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    Hi neverforgettovisitme

  2. neverforgetvivistee
    • 3 years ago
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    hi

  3. neverforgetvivistee
    • 3 years ago
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    lol

  4. moneybird
    • 3 years ago
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    6(+-)√53 / 2 is simplified

  5. Hero
    • 3 years ago
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    :D

  6. Tomas.A
    • 3 years ago
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    answer is is \[3\pm\sqrt{13}\]

  7. neverforgetvivistee
    • 3 years ago
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    wait how is it simplified? in another problem there was a problem where you divided by 3 because it was a common factor

  8. neverforgetvivistee
    • 3 years ago
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    wait i'll go look for it

  9. neverforgetvivistee
    • 3 years ago
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    -9x^2-9x+6

  10. satellite73
    • 3 years ago
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    no you cannot because the number inside the radical contains no perfect squares. this is the best you can do

  11. neverforgetvivistee
    • 3 years ago
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    later in the problem the guy got 9(+-)3√33 --------- -18 then he simplified by 3

  12. neverforgetvivistee
    • 3 years ago
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    khanacademy

  13. satellite73
    • 3 years ago
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    for \[-9x^2-9x+6=0\] you can start with \[3x^2+3x-2=0\] and reduce before you start

  14. neverforgetvivistee
    • 3 years ago
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    oh i see

  15. satellite73
    • 3 years ago
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    hold the phone solution to \[ x^2-6x-4=0\]is not what you wrote. it is what tomas A wrote

  16. Tomas.A
    • 3 years ago
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    \[\frac{9\pm3\sqrt{33}}{-18}=\] you can simplify this since \[\frac{3(3\pm\sqrt{33})}{-6\cdot3}=\] \[\frac{1(3\pm\sqrt{33})}{-6\cdot1}=\]

  17. satellite73
    • 3 years ago
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    \[x^2-6x-4=0\] \[x^2-6x=4\] \[(x-3)^2=4+9=13\] \[x-3=\pm\sqrt{13}\] \[x=3\pm\sqrt{13}\]

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