1. ## Help Solving a Quadratic

I am having trouble figuring out this problem. I know that there may not be a real solution, but we have to show the steps to solving this theoretical quadratic. If someone could show me the steps that would be awesome!

3x^2 + xy + 2y^2 = 0

2. Originally Posted by KarlosK
I am having trouble figuring out this problem. I know that there may not be a real solution, but we have to show the steps to solving this theoretical quadratic. If someone could show me the steps that would be awesome!

3x^2 + xy + 2y^2 = 0
Is it a quadratic in x or y?

3. I don't really understand your question if it is in x or y? She just asked for us to solve this using the quadratic equation. I assume that there is going to be no number answer

4. Originally Posted by KarlosK
I don't really understand your question if it is in x or y? She just asked for us to solve this using the quadratic equation. I assume that there is going to be no number answer
If it's a quadratic in y you need to solve for y=f(x)
If it's a quadratic in x you need to solve for x = f(y)

5. Originally Posted by KarlosK
I don't really understand your question if it is in x or y? She just asked for us to solve this using the quadratic equation. I assume that there is going to be no number answer
If you're solving for $\displaystyle x$ and using the quadratic formula: $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ let $\displaystyle a=3,\ b=y$ and $\displaystyle c=2y^2$

Spoiler:
$\displaystyle x=\frac{-y\pm\sqrt{-23y^2}}{6}$

6. What you did there makes sense. I understand how to get to that point now. So basically she is just looking for how I would come to the solution... but there will not be an actual number solution?

7. There are no real solutions, if that is what you are referring to, as $\displaystyle y^2$ will be positive, and you cant take the square root of a negative value.

8. What expression do you get inside the square root?

Is there any value of $\displaystyle y$ for which this expression will not be negative?