`4x^4+15x^2=4`, so

`4x^4+15x^2-4=0`

Let `u=x^2` to make things easier. Then we have:

`4u^2+15u-4=0`

`4u^2+16u-u-4=0`

`4u(u+4)-1(u+4)=0`

`(4u-1)(u+4)=0`

So `u=1/4` or `u=-4`.

But `u=x^2`, so

`x^2=1/4` or `x^2=-4`

`x=+-1/2` are the only real solutions.

Once again, let's have a look at the graph of the function, to better understand the situation. This time it's `y = 4x^4 + 15x^2 - 4`.

The intersection with the x-axis will tell us the solution for the original equation.

0.51-0.5-151015-5xyOpen image in a new page

Graph of `y = 4x^4 + 15x^2 - 4`

The two intersections with the x-axis are at `x = -0.5` and `x=0.5`.

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