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cos00155079

  • 2 years ago

p^2x^2+2px+1=0 solve using the quadratic formula

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  1. cos00155079
    • 2 years ago
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    |dw:1335056613189:dw| what is a? or how do you solve for a here?

  2. bmp
    • 2 years ago
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    Assuming p is a real number, p^2 = a, 2p = b and c = 1, because they are all constants.

  3. cos00155079
    • 2 years ago
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    so what is the rest...? do i just ignore it? the xs?

  4. bmp
    • 2 years ago
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    x is the x in the quadratic equation. It will become:\[x = (-2p \pm \sqrt{(2p)^{2} - 4p^{2}})/(2p^{2})\]

  5. imagreencat
    • 2 years ago
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    Actually, it's just (px)^2 + 2px + 1 = 0. We can just easily factor this especially since C here only has (1)(1) as its factors. (px + 1)(px + 1). :)

  6. exraven
    • 2 years ago
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    a quadratic equation can be written in the form:\[ax^2 + bx + c = 0\]then we get\[a = p^2\]\[b = 2p\]\[c = 1\]after that you just need to plug them to the quadratic formula\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

  7. bmp
    • 2 years ago
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    @imagreencat Indeed, that's easier to solve :-)

  8. cos00155079
    • 2 years ago
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    thanks everyone :)

  9. imagreencat
    • 2 years ago
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    You're welcome!

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