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Quadratic Inequalities

  1. Dec 25, 2005 #1
    (a) If the roots of the equation 2(x)^2 + kx + 100 = 0 are positive,
    find the possible range of k.
    (b) If, in addition, one root is twice the other, find the roots and the value of k.

    I have tried (a), but incorrect:
    discriminate > 0
    k^2 - (4)(2)(100) > 0
    k^2 > 800
    k > + or - 20(2)^1/2
    What's wrong with my calculation?
    The correct ans is:k is less than or equal to - 20 (2)^1/2
    2. Find the values of t for which the quadratic equation
    (x)^2 - tx + t + 3 = 0 has one positive root and one negative root.

    >>> I have no idea to start doing it.
    The correct answer is t < -3
    Can anyone help? :blushing:
    Wish you all have a Merry Christmas! :rofl:
  2. jcsd
  3. Dec 25, 2005 #2


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    For #1, since the problem does not ask for two distinct real roots, the discriminant can also be 0, ie: [itex]\Delta \geq 0[/itex].
    Solving that, you have:
    [tex]\left[ \begin{array}{l} k \geq 20 \sqrt{2} \\ k \leq -20 \sqrt{2} \end{array} \right. \ (1)[/tex] (Is this what you get?)
    But since the root(s) must also be positive (let x1, and x2 be the roots of the equation), you must also have:
    [tex]\left\{ \begin{array}{l} x_1 + x_2 > 0 \\ x_1 x_2 > 0 \end{array} \right.[/tex].
    Using Vi├Ęte's formulas, we have:
    [tex]\left\{ \begin{array}{l} -\frac{k}{2} > 0 \\ 50 > 0 \end{array} \right. \ (2)[/tex].
    So for what k, does that equation have positive root(s)?
    If one root is twice the other, assume that x2 = 2 x1.
    Then your quadratic equation must be in some form of:
    [tex]\alpha(x - x_1) (x - x_2) = \alpha(x - x_1) (x - 2x_1) = \alpha x ^ 2 - 3 \alpha x_1 x + 2 \alpha x_1 ^ 2 = 0, \ \mbox{where } \alpha \mbox{ is some number.}[/tex]
    That means: [tex]2x ^ 2 + kx + 100 = \alpha x ^ 2 - 3 \alpha x_1 x + 2 \alpha x_1 ^ 2[/tex], so what's [tex]\alpha , \ x_1, \ x_2, \ k[/tex]?
    #2, If one root is positive, and one is negative then x1 x2 < 0, right? That means:
    [tex]t + 3 < 0[/tex].
    Remember that when we have: [tex]x_1 x_2 = \frac{c}{a} < 0 \Leftrightarrow ca < 0[/tex], we also have: [tex]\Delta = b ^ 2 - 4ac > 0[/tex], that means the equation must also have 2 roots. So here we don't need to find the k value for which the discriminant is greater than 0 (since it's already greater than 0). We just have to solve:
    [tex]x_1 x_2 = \frac{c}{a} < 0[/tex] for k.
    Can you go from here?
    Last edited: Dec 25, 2005
  4. Dec 25, 2005 #3


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    Staff Emeritus
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    Gold Member

    This one's easy: the question asks you to find a range for k when you know the roots are positive. Your derivation doesn't involve the roots at all! (Let alone involve the fact the roots are positive) Thus, it should be no surprise that you got the wrong answer.

    For almost every mathematical problem (at least for every homework problem), there is an obvious way to start. It may seem like a trivial step, but it is very frequently useful.

    That method is to simply rewrite the problem in terms of the definitions.

    You're talking about two things: a positive root, and a negative root. So, you should give them names! (I'll use a and b)
    Then, you should write down the formulas that say that a is a positive root of that equation. (I.E. the formula that says a is a root, and the formula that says a is positive)
    Then, you do the same thing for b.

    It might not always be the best place to start, but you should always be able to start the problem. (e.g. VietDao suggests using a particular theorem about the roots of a polynomial, rather than just starting with the formulas that say the roots really are roots)
    Last edited: Dec 25, 2005
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