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Thread: 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 $\displaystyle x = p \pm \sqrt{p^2 - p + 2}$.

    So you require the values of p that satisfy the inequality $\displaystyle p < \sqrt{p^2 - p + 2}$ (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 $\displaystyle p < \sqrt{p^2 - p + 2}$ (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 $\displaystyle ax^2 + bx + c = 0$ is $\displaystyle \frac{c}{a}$

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

    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 $\displaystyle x^2 - 2px + p - 2 = 0$ has 2 real roots with one positive and one negative if $\displaystyle p-2 < 0$ then $\displaystyle p < 2$
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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 $\displaystyle x = p \pm D$, that is, $\displaystyle x = p + D$ and $\displaystyle x = p - D$ where $\displaystyle D = \sqrt{p^2 - p + 2}$.

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

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