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  1. #1
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    quadratic

    The quadratic equation 2x^2 + (p+1)x + 8 = 0 find possible values of p.

    thanks
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by gracey View Post
    The quadratic equation 2x^2 + (p+1)x + 8 = 0 find possible values of p.

    thanks
    Possible values based on what criteria? As it stands p could have any value.

    -Dan
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  3. #3
    GAMMA Mathematics
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    Quote Originally Posted by gracey View Post
    The quadratic equation 2x^2 + (p+1)x + 8 = 0 find possible values of p.

    thanks
    Use the quadratic equation.

    x=\frac{-(p+1) \pm \sqrt{(p+1)^2 - 4(8)(2)}}{4}

    p+1 is restricted by the radical or discriminant in order for x to have real solutions.

    (p+1)^2 - 4(8)(2) \ge 0

    p^2 + 2p -63 \ge 0

    (p+9)(p-7) = 0

    p = -9, 7

    You have to test the three values that are less than -9, between -9 and 7, and greater than 7 to see which sub domains hold for the inequality.

    It turns out that p \le 9, p \ge 7 OR p: (-\infty, -9] \cap [7, \infty)
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    Forum Admin topsquark's Avatar
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    Ah but the given condition is apparently nothing more than that the equation be valid. No one said we can't have complex zeros. I was waiting for verification on that point.

    -Dan
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  5. #5
    GAMMA Mathematics
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    Quote Originally Posted by topsquark View Post
    Ah but the given condition is apparently nothing more than that the equation be valid. No one said we can't have complex zeros. I was waiting for verification on that point.

    -Dan
    True, but when is a problem stated with all the given conditions. A lot is assumed in elementary problems, let alone all mathematical problems outside of a full proof.
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  6. #6
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    thankyou everyone
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