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anonymous

  • one year ago

Am I right?

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  1. anonymous
    • one year ago
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    Find the vertex of f(x) = 5x2 + 2x + 1. (-2±√(2^2-4*5*1))/(2*5) (-2±√(-16))/10 = (-2±-4)/10 = 2/10 and -6/10

  2. campbell_st
    • one year ago
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    no you are incorrect.... it would seem that you have found the x intercepts or the zeros for the quadratic

  3. anonymous
    • one year ago
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    Oh, I mixed up the problem, that is what I was trying to find, LOL. Sorry, working all day on Algebra.

  4. anonymous
    • one year ago
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    I was trying to find the zeroes. Is it right?

  5. campbell_st
    • one year ago
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    if this is algebra the 1st step is to find the line of symmetry for the parabola, as the vertex is on the line of symmetry so for a parabola \[y=ax^2 + bx + c\] the line of symmetry is \[x = \frac{-b}{2 \times a}\] you have b = 2 and a = 5 when you get the value for x, substitute the x value into the equation to find the minimum value these to values will get the vertex... as a point

  6. phi
    • one year ago
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    2/10 and -6/10 you probably should simplify these to 1/5 and -3/5

  7. anonymous
    • one year ago
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    Yes, I did. So it's right?

  8. phi
    • one year ago
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    oh, one subtlety! \[ \frac{-2 \pm \sqrt{-16}} {10 } \]

  9. phi
    • one year ago
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    the square root of -16 is sqrt(16) * sqrt(-1) or 4 i (we use i for the square root of -1)

  10. anonymous
    • one year ago
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    So, I put i after the top negative number?

  11. phi
    • one year ago
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    one root is \[ \frac{-1}{5} + \frac{2}{5} i \] the other is \[ \frac{-1}{5} - \frac{2}{5} i \]

  12. anonymous
    • one year ago
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    You mean 3/5?

  13. phi
    • one year ago
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    \[\frac{-2 \pm \sqrt{-16}} {10 } \\ \frac{-2 \pm 4i }{10 }\\ \frac{-2}{10} \pm \frac{4}{10} i \]

  14. phi
    • one year ago
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    and finally \[ - \frac{1}{5} \pm \frac{2}{5} i \]

  15. phi
    • one year ago
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    you cannot combine the term with "i" with the "real" term the term with "i" is called the imaginary part and the other part is called the "real part"

  16. anonymous
    • one year ago
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    But, isn't the square root of -16, -4?

  17. phi
    • one year ago
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    -4*-4 = 16 (which is obviously not -16) people were very confused about this a few hundred years ago. they finally decided the only way to get -16 was to use \( \sqrt{-1} \) and then \[ \sqrt{-1} \cdot 4 \cdot \sqrt{-1} 4 = \sqrt{-1} \cdot \sqrt{-1} 16 = -1\cdot 16=-16\]

  18. anonymous
    • one year ago
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    Ok, sorry, not very good at math, I'm a history and English guy.

  19. phi
    • one year ago
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    just remember \( \sqrt{-16} = 4i \)

  20. anonymous
    • one year ago
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    Ok, thanks!!!

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