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anonymous

  • 4 years ago

find a, b, c if root of equation y=ax^2+bx+c root is (1/5)-(2/9)i i means imaginary

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  1. Mertsj
    • 4 years ago
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    If 1/5 -(2/9)i is a root then 1/5+(2/9)i is also a root. So write that [x-(1/5-(2/9)i][x-1/5+(2/9)i]=0 and simplify into the ax^2+bx+c form

  2. anonymous
    • 4 years ago
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    Thanks but I'm having trouble multiplying it, but I understand what you mean, like the conjugates multiply to get the equation.

  3. anonymous
    • 4 years ago
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    Also, aren't you supposed to switch the + and - in the second part? [x-(1/5-(2/9)i][x+1/5-(2/9)i]=0

  4. Mertsj
    • 4 years ago
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    \[[x-(\frac{1}{5}-\frac{2}{9}i)][x-(\frac{1}{5}+\frac{2}{9}i)]=0\]

  5. Mertsj
    • 4 years ago
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    \[x ^{2}-(\frac{1}{5}+\frac{2}{9}i)x-(\frac{1}{5}-\frac{2}{9}i)x+(\frac{1}{5}-\frac{2}{9}i)(\frac{1}{5}+\frac{2}{9}i)\]

  6. Mertsj
    • 4 years ago
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    \[x ^{2}-\frac{1}{5}x-\frac{2}{9}ix-\frac{1}{5}x+\frac{2}{9}ix+\frac{1}{25}+\frac{4}{81}\]

  7. Mertsj
    • 4 years ago
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    \[x ^{2}-\frac{2}{5}x+\frac{181}{2025}=0\]

  8. Mertsj
    • 4 years ago
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    \[2025x ^{2}-810x+181=0\]

  9. Mertsj
    • 4 years ago
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    a= 2025, b=-810, c=181

  10. Mertsj
    • 4 years ago
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    Are you still with me?

  11. anonymous
    • 4 years ago
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    Yup

  12. Mertsj
    • 4 years ago
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    Did I made any more mistakes?

  13. anonymous
    • 4 years ago
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    That was a masterpiece

  14. Mertsj
    • 4 years ago
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    ty

  15. anonymous
    • 4 years ago
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    no i think you did it perfectly

  16. anonymous
    • 4 years ago
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    Thanks

  17. Mertsj
    • 4 years ago
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    yw

  18. anonymous
    • 4 years ago
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    I just checked it and it worked :)

  19. Mertsj
    • 4 years ago
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    Great.

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