# What do I have to solve in the following? If complex, how do I solve it?

x^4 + x^2 -2 = 0

I broke my head on the topic. Please help..

- The only instruction provided for me was to 'solve'

### 4 Answers

- 8 years agoBest Answer
Use a "u" substitution: u = x^2. Then the quadraticity becomes apparent as you rewrite as u^2 + u - 2 = 0. Now we're looking for factors of -2 adding to 1. If you said 2 and -1, you're right on: (u+2)(u-1)=0, and solve using the standard method from there to solve for u, and then converting back to x.

To reinforce your understanding, I've searched and found a webpage and a video tutorial that discuss how to address problems similar to this one that gave you the trouble. I thought sharing these might be helpful to you. I've listed them below.

As always, if you need more help, please clarify where you are in the process and what's giving you trouble. I'd be more than happy to continue to assist.

Source(s): http://library.thinkquest.org/20991/alg2/quad.html http://www.youtube.com/watch?v=MkluknSI4wU - 8 years ago
To break this down simply, say the equation was as follows:

x^2 + x - 2 = 0 (similar equation as the one above just with smaller power)

the equation would factor out to having roots of:

(x + 2)(x - 1)

This is similar to the one you listed as the x components would just be squared:

(x^2 + 2)(x^2 -1)

You can always check your solution but re-solving the roots and seeing if it comes back to the given equation, if not you know that you did something wrong and need to look back through your work.

Source(s): High School Math - PolygonLv 78 years ago
First substitute y = x^2

equation then becomes y^2 + y - 2 = 0

(y +2)(y - 1) = 0

so y = -2 or + 1

so x^2 = -2 or 1

x = i(Square root 2) or + or - 1

- wirehawkbostonLv 78 years ago
x^4 + x^2 -2 = 0

x^4 + x^2 = 2

x/^2(x^2 + 1) = 2

1. x^2 = 2

x = sq rt of 2

2 & 3. x^2 + 1 + 0

x^2 = sq rt of 1

x = +/- 1