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anonymous

  • one year ago

How do I do 7x^2 + x + 2 = 0 by the quadratic formula. Thanks for your help in advance!

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  1. DanJS
    • one year ago
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    do you know the quadratic formula?

  2. anonymous
    • one year ago
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    I think so

  3. DanJS
    • one year ago
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    ax^2 + bx + c = 0 7x^2 + 1x + 2 = 0

  4. DanJS
    • one year ago
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    use those values for a, b , and c

  5. anonymous
    • one year ago
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    a = 7 b = 1 c = 2?

  6. DanJS
    • one year ago
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    yep.. \[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

  7. anonymous
    • one year ago
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    |dw:1441589861398:dw|

  8. DanJS
    • one year ago
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    the stuff under the square root turns out to be negative, so no real solutions to this one, less you want the complex roots

  9. DanJS
    • one year ago
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    \[\sqrt{-1}=i\]

  10. anonymous
    • one year ago
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    I do not know how to solve this from start to finish.

  11. DanJS
    • one year ago
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    you have it right so far with the drawing, just have to simplify it down now if needed.

  12. anonymous
    • one year ago
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    I tried it but when it got down to the imaginary number it got confusing.

  13. DanJS
    • one year ago
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    \[x=\frac{ -1 \pm \sqrt{-55}}{ 14 }\] \[x=\frac{ -1 }{ 14 }\pm \frac{ \sqrt{55} *\sqrt{-1}}{ 14 }\]

  14. anonymous
    • one year ago
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    |dw:1441590203726:dw|

  15. DanJS
    • one year ago
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    just have to remember...\[i=\sqrt{-1}\] put in i for that then solve normally

  16. DanJS
    • one year ago
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    \[x =\frac{ -1 }{ 14 }+\frac{ \sqrt{55} }{ 14 }*i ~~~~~or~~~~~x =\frac{ -1 }{ 14 }-\frac{ \sqrt{55} }{ 14 }*i\]

  17. anonymous
    • one year ago
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    Oh thank you I never knew what a imaginary number was.

  18. anonymous
    • one year ago
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    Does it simplified more is that the final answer?

  19. DanJS
    • one year ago
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    welcome, If you have to know nothing else about imaginary numbers, then just remember \[i^2=-1~~~~or~~~~~i=\sqrt{-1}\]

  20. DanJS
    • one year ago
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    that is all you can do to make it more simple looking

  21. anonymous
    • one year ago
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    Thanks so much! Did the i get in the problem because square root of 55 is a decimal?

  22. DanJS
    • one year ago
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    no, because the quadratic formula turned out to have a negative number in the square root, remember you cant take the square root of a number smaller than zero

  23. DanJS
    • one year ago
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    b^2 - 4*a*c was -55

  24. anonymous
    • one year ago
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    Oh okay I see how it got in then. Thanks for your time and help!

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