# greater than, equal to or less than zero

• September 20th 2008, 08:39 AM
georgel
greater than, equal to or less than zero
I'm not sure this is the right forum, but here goes:

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

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

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

I've tried and tried, but no luck.

• September 20th 2008, 08:53 AM
Jhevon
Quote:

Originally Posted by georgel
I'm not sure this is the right forum, but here goes:

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

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

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

I've tried and tried, but no luck.

good job so far

now, as for the latter part

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

plot the solutions on a number line and test the intervals. you should be able to determine the range of values for $\alpha$ that causes each of the conditions you are considering
• September 20th 2008, 08:57 AM
bkarpuz
Quote:

Originally Posted by georgel
I'm not sure this is the right forum, but here goes:

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

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

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

I've tried and tried, but no luck.

you are sure in the right place and on the right direction.
$y^{2}(x^{2}+x+\alpha)=y^{2}\bigg(x+\frac{1-\sqrt{1-4\alpha}}{2}\bigg)\bigg(x+\frac{1+\sqrt{1-4\alpha}}{2}\bigg)$
I hope it is now clear... (Wink)
• September 20th 2008, 09:00 AM
skeeter
$x^2 + x + \alpha = 0$

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

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

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

if $1 - 4\alpha > 0$ , then $x^2 + x + \alpha \leq 0$ for $\frac{-1 - \sqrt{1 - 4\alpha}}{2} \leq x \leq \frac{-1 + \sqrt{1 - 4\alpha}}{2}$
• September 23rd 2008, 04:09 AM
georgel
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.