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dddan

  • 2 years ago

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

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

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

  3. dddan
    • 2 years ago
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    ya but i got 4+- the square root of -24 / 4 does this make sense?

  4. hartnn
    • 2 years ago
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    your b=-5, how did u get 4+/- ... ?

  5. dddan
    • 2 years ago
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    sry the b=-4 and the c=5

  6. wio
    • 2 years ago
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    You have imaginary roots. Is that what you're worried about?

  7. dddan
    • 2 years ago
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    yup

  8. wio
    • 2 years ago
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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

  9. hartnn
    • 2 years ago
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    u actually have imaginary roots for this..

  10. dddan
    • 2 years ago
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    alright can u tell me where to take it from the imaginary roots?

  11. hartnn
    • 2 years ago
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    how did u get square root of -24 ?

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

  13. hartnn
    • 2 years ago
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    lol! b=-5 not -4

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

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

  16. dddan
    • 2 years ago
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    alright thank you very much!

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