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anonymous

  • 5 years ago

need help one more time on solving quadratic equation

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  1. anonymous
    • 5 years ago
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    ok

  2. anonymous
    • 5 years ago
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    \[y ^{2}=-24y+144\] and so far I have \[-3 + or -\sqrt{57}\over -12\]

  3. anonymous
    • 5 years ago
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    I'm sure it's supposed to simplify from there but not sure exactly how. Or if this will give me 2 non real answers, or 2 real ones

  4. anonymous
    • 5 years ago
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    3 and 19 make 57

  5. anonymous
    • 5 years ago
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    \[-24\pm \sqrt{b ^{}}\]

  6. anonymous
    • 5 years ago
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    I got 0 under radical so -24+ -0/2

  7. anonymous
    • 5 years ago
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    -12

  8. anonymous
    • 5 years ago
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    ok, so -24 + and - -12 are the non real answers?

  9. anonymous
    • 5 years ago
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    wait! I just realized I typed the wrong problem..

  10. anonymous
    • 5 years ago
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    I used +144 not -144 sorry my oversight

  11. anonymous
    • 5 years ago
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    Sorryy, can you guys wait a sec while I put the right one up here

  12. anonymous
    • 5 years ago
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    Mary, when you're solving these, be sure to get them in the form \[ax^2 + bx + c = 0\] before plugging into the quadratic equation

  13. anonymous
    • 5 years ago
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    \[-6^{2}+3x+2=0\]

  14. anonymous
    • 5 years ago
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    So what is a, b, and c?

  15. anonymous
    • 5 years ago
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    b is 3 and c is 2

  16. anonymous
    • 5 years ago
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    I'm guessing that first term should be \(-6x^2\)

  17. anonymous
    • 5 years ago
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    so a is?

  18. anonymous
    • 5 years ago
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    oh yeah, sorry i thought you just asked about b and c

  19. anonymous
    • 5 years ago
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    a is -6

  20. anonymous
    • 5 years ago
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    Ok, now what do you get when you plug them in?

  21. anonymous
    • 5 years ago
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    the formula we have been going off of is a bit different than that. But I had gotten -3 + and - root 57 over -12

  22. anonymous
    • 5 years ago
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    \[x=\frac{-3 \pm \sqrt{9-4(-6)(2)}}{2(-6)}\] \[x = \frac{-3 \pm \sqrt{9+48}}{-12}\] \[x = \frac{3 \pm \sqrt{57}}{12}\]

  23. anonymous
    • 5 years ago
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    81 under radical bc of double neg so you add

  24. anonymous
    • 5 years ago
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    sorry ya'll numbers all wrong I will leave

  25. anonymous
    • 5 years ago
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    don't forget -3

  26. anonymous
    • 5 years ago
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    ok so I had the right answer, except it's positive 3, not negative?

  27. anonymous
    • 5 years ago
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    are they real or non real solutions?

  28. anonymous
    • 5 years ago
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    You can simplify a bit further, but it's not gonna be a whole number or even a rational one. \[x = \frac{3 \pm \sqrt{57}}{12}\] \[x = \frac{3}{12} \pm \frac{\sqrt{57}}{12}\] \[x = \frac{1}{4} \pm \frac{\sqrt{57}}{12}\]

  29. anonymous
    • 5 years ago
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    They are real solutions. You only get non-real solutions when the thing under the square root is negative.

  30. anonymous
    • 5 years ago
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    As for the -3, if you factor a -1 from the top and bottom you can cancel to get a positive 3 on top and a 12 on bottom. So it's going to be positive. It doesn't matter much though because one of your x will be positive and one is negative.

  31. anonymous
    • 5 years ago
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    Your answer was also correct, but could be simplified a little bit.

  32. anonymous
    • 5 years ago
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    ok, thanks!

  33. anonymous
    • 5 years ago
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    because the - was common to both the numerator and denominator.

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