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

Jul 4, 2015

$0 = 2 {x}^{4} + {x}^{2} - 3$

$= \left(2 {x}^{2} + 3\right) \left({x}^{2} - 1\right)$

$= \left(2 {x}^{2} + 3\right) \left(x - 1\right) \left(x + 1\right)$

So $x = 1$ or $x = - 1$

#### Explanation:

$2 {x}^{4} + {x}^{2} - 3 = 2 {\left({x}^{2}\right)}^{2} + \left({x}^{2}\right) - 3$

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 $\left({x}^{2} - 1\right)$ is a factor.

${x}^{2} - 1 = {x}^{2} - {1}^{2} = \left(x - 1\right) \left(x + 1\right)$

using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

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

$2 {x}^{2} + 3 = 0$ has no real roots, since ${x}^{2} \ge 0$, hence $2 {x}^{2} + 3 \ge 3$

So the only real roots of the original quartic equation are $x = \pm 1$