# Equations Quadratic in Form ;_;

1. Dec 25, 2011

### Matthewkind

1. The problem statement, all variables and given/known data
y^4 + 3y^2 - 4 = 0

2. Relevant equations

3. The attempt at a solution
The base, I know, is y^2. So in order to make this a quadratic equation, we come up with an arbitrary variable, say α, which is equal to the base. Re-writing this, we get:

(y^2)^2 + 3(y^2) - 4 = 0
α = y^2
α^2 + 3α - 4 = 0

So now it's a quadratic equation.
Factoring, we get:

(α - 1) (α + 4) = 0

Setting both of these equal to zero, we get:

α = 1 and α = -4

This next part is where I'm lost, so maybe one of you bright people can show me where I'm making my mistake.

Since α = y^2, -4 must = y^2 too.

-4 = y^2

Solving this for y, we get:

y = sqrt(-4) = 2i.

However, in the text - without any explanation as to how this occurs (which is what is confusing me about these problems) - it says that the two solutions are 2i and -2i.

Can someone please explain this step to me. D:

2. Dec 25, 2011

### Matthewkind

Oh. Wait. I think I figured it out. When you take the square root of something we have to remember to put plus/minus, right? ;_; Maybe it's time to go back to radical rules! How embarrassing! Sorry, Physics Forums!

3. Dec 25, 2011

### mtayab1994

Yep that's exactly what it is you can have to answers ±√.