Solving the equation $2x^4-3x^2-9=0$

Question

How would I go about solving $2x^4-3x^2-9=0$?

I started by taking $x^2$ out, thus getting $x^2(2x^2-3) = 9$ but I don't know if that did anything useful, usually if this equation is equal to $0$ it's easy but this time it's equal to $9$ so I'm lost.

Answer 1

The standard trick here is to write $X=x^2$ and solve $2X^2-3X-9=0$.

Answer 2

You first view $x^2$ as one variable: let $y=x^2$, then solve for $2y^2-3y-9=0$, which has two solutions, with one of them $y_0$ positive. Since $y=x^2>0$ (assuming $x$ is real), we have to ignore the negative root. So $x=\pm \sqrt{y_0}$.

Answer 3

Fortunately the polynomial can be factored into $$ (2x^2 + 3)(x^2 - 3)=0. $$ This again allows to use the formula for quadratic equations.

Answer 4

Suppose,$y$=$x^2$. So, the equation becomes $2y^2-3y-9=0$. Using Sridhar Acharya's theorem we get $y=3,-\frac{3}{2}$. So finally,$x=\sqrt{3}$ (as square root of a negative numbar is not possible) .

Answer 5

We can add $3x^2-3x^2$ to the left-hand side and factor by grouping.$$\begin{align*}2x^4-6x^2+3x^2-9=0 & \iff 2x^2(x^2-3)+3(x^2-3)=0\\ & \iff (x^2-3)(2x^3+3)=0\end{align*}$$However, if you did not see that sneaky trick, we can always make the substitution $X=x^2$ to get your polynomial into a quadratic.$$2X^2-3X-9=0\iff (X-3)(2X+3)=0\iff (x^2-3)(2x^3+3)=0$$In fact, fourth degree polynomials of the form $ax^4+bx^2+c$ are called biquadratics.