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dddan Group Title

if a=2 ; b=-5; c=5 how would you solve the quadratic equation?

  • one year ago
  • one year ago

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  1. hartnn Group Title
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    then the two roots of x are: \(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

    • one year ago
  2. wio Group Title
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    Use the quadratic foumula \[x_1 =\frac{ -b + \sqrt{b^2-4ac}} {2a}\] \[x_2 =\frac{ -b - \sqrt{b^2-4ac}} {2a}\]

    • one year ago
  3. dddan Group Title
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    ya but i got 4+- the square root of -24 / 4 does this make sense?

    • one year ago
  4. hartnn Group Title
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    your b=-5, how did u get 4+/- ... ?

    • one year ago
  5. dddan Group Title
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    sry the b=-4 and the c=5

    • one year ago
  6. wio Group Title
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    You have imaginary roots. Is that what you're worried about?

    • one year ago
  7. dddan Group Title
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    yup

    • one year ago
  8. wio Group Title
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    Why? Is it a physics problem and you are supposed to have real roots? Because there is nothing wrong with the quadratic formula. Double check your coefficients

    • one year ago
  9. hartnn Group Title
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    u actually have imaginary roots for this..

    • one year ago
  10. dddan Group Title
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    alright can u tell me where to take it from the imaginary roots?

    • one year ago
  11. hartnn Group Title
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    how did u get square root of -24 ?

    • one year ago
  12. dddan Group Title
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    because with the coefficients, its (-4)^2-4(2)(5) , in the square root right?

    • one year ago
  13. hartnn Group Title
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    lol! b=-5 not -4

    • one year ago
  14. dddan Group Title
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    no i i mixed up the coefficents in the original question. i was given a=2;b=-4;c=5

    • one year ago
  15. hartnn Group Title
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    ok, \(\huge{x=\frac{4 \pm \sqrt{-24}}{4}}=\frac{4 \pm 2\sqrt{-6}}{4}=\frac{2\pm i\sqrt{6}}{2}\)

    • one year ago
  16. dddan Group Title
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    alright thank you very much!

    • one year ago
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