biquadratic equation

A biquadratic equation (in a narrower sense) is the special case of the quartic equation (http://planetmath.org/QuarticFormula) containing no odd degree terms:

$a{x}^4+b{x}^2+c=0$(1)

Here, $a$, $b$, $c$ are known real or complex numbers and  $a\ne 0$.

For solving a biquadratic equation (1) one does not need the quartic formula (http://planetmath.org/QuarticFormula) since the equation may be thought a quadratic equation with respect to $x^2$, i.e.

$$a{(x^2)}^2+b{x}^2+c=0,$$

whence

$$x^2=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

(see quadratic formula or quadratic equation in $ℂ$ (http://planetmath.org/QuadraticEquationInMathbbC)).  Taking square roots of the values of $x^2$ (see taking square root algebraically), one obtains the four roots (http://planetmath.org/Equation) of (1).

Example.  Solve the biquadratic equation

$x^4+x^2-20=0.$(2)

We have

$x^2={\displaystyle \frac{-1\pm \sqrt{1^2-4\cdot 1\cdot (-20)}}{2\cdot 1}}={\displaystyle \frac{-1\pm 9}{2}},$(3)

i.e.  $x^2=4$  or  $x^2=-5$.  The solution is

$x=\pm 2\mathit{\hspace{1em}}\vee \mathit{\hspace{1em}}x=\pm i\sqrt{5}.$(4)

Remark.  In one wants to form of rational numbers a polynomial equation with rational coefficients and most possibly low degree by using two square root operations, then one gets always a biquadratic equation.  A couple of examples:

1) $x=1+\sqrt{2}+\sqrt{3}$
${(x-1)}^2=2+2\sqrt{6}+3$
$y^2-5=2\sqrt{6}$
$y^4-10y^2+1=0$  (one has substituted (http://planetmath.org/TchirnhausTransformations)  $x-1:=y$)

2) $x=\sqrt{\sqrt{2}-1}$
$x^2=\sqrt{2}-1$
${(x^2+1)}^2=2$
$x^4+2x^2-1=0$