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Math Help - Discriminants

  1. #1
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    Exclamation Discriminants

    Hi

    For the question: Find all values of p so that x^2 - 2px + p - 2 = 0 has one positive root and a negative root.

    What is the condition of the discriminant for a positive and a negative root?

    Thanx
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    Hi

    For the question: Find all values of p so that x^2 - 2px + p - 2 = 0 has one positive root and a negative root.

    What is the condition of the discriminant for a positive and a negative root?

    Thanx
    Yoo should be able to use the quadratic formula to get x = p \pm \sqrt{p^2 - p + 2}.

    So you require the values of p that satisfy the inequality p < \sqrt{p^2 - p + 2} (why?)
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    So you require the values of p that satisfy the inequality p < \sqrt{p^2 - p + 2} (why?)
    Hmmmm I have no clue. Why exactly does it have to satisfy that specific inequality?
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  4. #4
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    Hi
    Maybe I can add something

    The product of the roots (real or complex) of equation ax^2 + bx + c = 0 is \frac{c}{a}

    Therefore ax^2 + bx + c = 0 has 2 real roots with one positive and one negative if both conditions are realised :
    (i) b^2 - 4ac \geq 0
    (ii) \frac{c}{a} < 0

    Condition (ii) implies that a and c have different sign ; then their product is negative
    It means that condition (ii) implies condition (i) and therefore only condition (ii) is sufficient

    In your specific case x^2 - 2px + p - 2 = 0 has 2 real roots with one positive and one negative if p-2 < 0 then p < 2
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  5. #5
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    Quote Originally Posted by xwrathbringerx View Post
    Hmmmm I have no clue. Why exactly does it have to satisfy that specific inequality?
    Look at the solution for x. It has the form x = p \pm D, that is, x = p + D and x = p - D where D = \sqrt{p^2 - p + 2}.

    x = p + D is always positive so that's the positive solution.

    The negative solution therefore has to come from x = p - D. So you require p - D < 0 \Rightarrow p < D \Rightarrow p < \sqrt{p^2 - p + 2}.
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