Algebra II

Explain, without doing any calculations, why there cannot be a Real solution to the equation
p5 + 2x2 + 3px + 7 + 8 = -2:

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asked by Kiara
  1. I suspect typos. The function is littered with junk.

    p?
    +7+8?

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    posted by Steve
  2. I will assume p5 is meant to be p^5
    and 2x2 is meant to be 2x^2 , then

    p^5 + 2x^2 + 3px + 17 = 0
    2x^2 + 3px + p^5 + 17 = 0
    a quadratic with a = 2 , b = 3p and c = p^5 + 17

    for real roots:
    9p^2 - 4(2)(p^5 + 17) ≥ 0
    9p^2 - 8p^5 - 136 ≥ 0


    using Wolfram to solve, I got p < -1.69
    http://www.wolframalpha.com/input/?i=9p%5E2+-+8p%5E5+-+136+%3D+0

    check your typing, as long as p < -1.69 you will have 2 real solutions each time

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    posted by Reiny

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