# Thread: greater than, equal to or less than zero

1. ## 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.

2. Originally Posted by georgel
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.

good job so far

now, as for the latter part

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

plot the solutions on a number line and test the intervals. you should be able to determine the range of values for $\displaystyle \alpha$ that causes each of the conditions you are considering

3. Originally Posted by georgel
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.

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...

4. $\displaystyle x^2 + x + \alpha = 0$

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

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 $\displaystyle \frac{-1 - \sqrt{1 - 4\alpha}}{2} \leq x \leq \frac{-1 + \sqrt{1 - 4\alpha}}{2}$

5. 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.