Question

How do you solve the equation for y: `2x^4+x^2-3=0`?

Answer 1

`0 = 2x^4+x^2-3`

`= (2x^2+3)(x^2-1)`

`= (2x^2+3)(x-1)(x+1)`

So `x = 1` or `x = -1`

Explanation:

`2x^4+x^2-3 = 2(x^2)^2 + (x^2) - 3`

is quadratic in `x^2`

Notice that the sum of the coefficients is `0`

`2+1-3 = 0`

So `x^2 = 1` is a root and `(x^2 - 1)` is a factor.

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

using the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

The other factor of `2x^4+x^2-3` must be `(2x^2+3)` in order that when multiplied by `(x^2-1)` it results in the leading `2x^4` term and the trailing `-3` term.

`2x^2+3 = 0` has no real roots, since `x^2 >= 0`, hence `2x^2+3 >= 3`

So the only real roots of the original quartic equation are `x = +-1`