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anonymous

  • 5 years ago

Solve using the quadratic forumula. 3x^2 + x - 4 = 0

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  1. anonymous
    • 5 years ago
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    Do you know the quadratic formula? \[-b \pm \sqrt{b ^{2}-4ac}/2a\]

  2. anonymous
    • 5 years ago
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    Then it's just a matter of plugging in the values and simplifying.

  3. anonymous
    • 5 years ago
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    3x^2+4x-3x-4=0 (x-1)(3x+4)=0

  4. anonymous
    • 5 years ago
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    x=1,-4/3

  5. anonymous
    • 5 years ago
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    I understand how to substitute everything in, but from there I am unsure how to simplify -1 ± 12−4*3*4−−−−−−−√

  6. anonymous
    • 5 years ago
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    -1 +/- sqrt -48

  7. anonymous
    • 5 years ago
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    a=3,b=1, c=-4

  8. anonymous
    • 5 years ago
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    Work on the square root part first. \[\sqrt{1^{2}-4*3*-4}\]

  9. anonymous
    • 5 years ago
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    You missed a couple things in the sqrt. \[1^{2}=1\] 4*3*-4=-48 1-(-48)

  10. anonymous
    • 5 years ago
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    Get that? So it's \[\sqrt{1+48}=\sqrt{49}=7\] So, \[(1 \pm 7)/6\]

  11. anonymous
    • 5 years ago
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    Thank you very much! That makes much more sense now. Once again, thanks!

  12. anonymous
    • 5 years ago
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    np.

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