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Thread: greater than, equal to or less than zero

  1. #1
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    greater than, equal to or less than zero

    I'm not sure this is the right forum, but here goes:

    Depending on $\displaystyle \alpha \in \mathbb{R}$, find where the expression

    $\displaystyle (xy)^2 + (\alpha + x)y^2$ is greater than zero, less than zero and equal to zero.

    I suppose the first step is to write it as
    $\displaystyle y^2 (x^2+x+\alpha)$.
    So, if $\displaystyle y=0$, than the expression is zero, no matter what the $\displaystyle \alpha$ is.
    Also, if $\displaystyle y>0$ or $\displaystyle y<0$, $\displaystyle y^2>0$, so it comes down to whether $\displaystyle x^2+x+\alpha$ is 0, <0 or >0.

    I've tried and tried, but no luck.

    Please, help.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by georgel View Post
    I'm not sure this is the right forum, but here goes:

    Depending on $\displaystyle \alpha \in \mathbb{R}$, find where the expression

    $\displaystyle (xy)^2 + (\alpha + x)y^2$ is greater than zero, less than zero and equal to zero.

    I suppose the first step is to write it as
    $\displaystyle y^2 (x^2+x+\alpha)$.
    So, if $\displaystyle y=0$, than the expression is zero, no matter what the $\displaystyle \alpha$ is.
    Also, if $\displaystyle y>0$ or $\displaystyle y<0$, $\displaystyle y^2>0$, so it comes down to whether $\displaystyle x^2+x+\alpha$ is 0, <0 or >0.

    I've tried and tried, but no luck.

    Please, help.
    good job so far

    now, as for the latter part

    solve $\displaystyle x^2 + x + \alpha = 0$

    how? quadratic formula.

    plot the solutions on a number line and test the intervals. you should be able to determine the range of values for $\displaystyle \alpha$ that causes each of the conditions you are considering
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  3. #3
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by georgel View Post
    I'm not sure this is the right forum, but here goes:

    Depending on $\displaystyle \alpha \in \mathbb{R}$, find where the expression

    $\displaystyle (xy)^2 + (\alpha + x)y^2$ is greater than zero, less than zero and equal to zero.

    I suppose the first step is to write it as
    $\displaystyle y^2 (x^2+x+\alpha)$.
    So, if $\displaystyle y=0$, than the expression is zero, no matter what the $\displaystyle \alpha$ is.
    Also, if $\displaystyle y>0$ or $\displaystyle y<0$, $\displaystyle y^2>0$, so it comes down to whether $\displaystyle x^2+x+\alpha$ is 0, <0 or >0.

    I've tried and tried, but no luck.

    Please, help.
    you are sure in the right place and on the right direction.
    $\displaystyle y^{2}(x^{2}+x+\alpha)=y^{2}\bigg(x+\frac{1-\sqrt{1-4\alpha}}{2}\bigg)\bigg(x+\frac{1+\sqrt{1-4\alpha}}{2}\bigg)$
    I hope it is now clear...
    Last edited by bkarpuz; Sep 20th 2008 at 09:07 AM. Reason: Moo's remark is prompted.
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  4. #4
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    $\displaystyle x^2 + x + \alpha = 0$

    quadratic formula ...

    $\displaystyle x = \frac{-1 \pm \sqrt{1 - 4\alpha}}{2}$

    if $\displaystyle 1 - 4\alpha < 0$ , then the roots for x are imaginary and
    $\displaystyle x^2 + x + \alpha > 0$.

    if $\displaystyle 1 - 4\alpha = 0$ , then x has a single double root at $\displaystyle x = -\frac{1}{2}$ and
    $\displaystyle x^2 + x + \alpha \geq 0$.

    if $\displaystyle 1 - 4\alpha > 0$ , then $\displaystyle x^2 + x + \alpha \leq 0$ for $\displaystyle \frac{-1 - \sqrt{1 - 4\alpha}}{2} \leq x \leq \frac{-1 + \sqrt{1 - 4\alpha}}{2}$
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  5. #5
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    That's exactly what I did the first time, and got zero points. :-(
    I didn't know what else to try.

    Thanks to everyone, I'll ask my teaching assistant what seems to be the problem.
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