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# biquadratic equation

A biquadratic equation (in a narrower sense) is the special case of the quartic equation containing no odd degree terms:

$\displaystyle ax^{4}+bx^{2}+c=0$ | (1) |

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

For solving a biquadratic equation (1) one does not need the quartic formula since the equation may be thought a quadratic equation with respect to $x^{2}$, i.e.

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

whence

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

(see quadratic formula or quadratic equation in $\mathbb{C}$). Taking square roots of the values of $x^{2}$ (see taking square root algebraically), one obtains the four roots of (1).

Example. Solve the biquadratic equation

$\displaystyle x^{4}+x^{2}-20=0.$ | (2) |

We have

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

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

$\displaystyle x=\pm 2\quad\lor\quad x=\pm i\sqrt{5}.$ | (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 $x-1:=y$)

2) $x=\sqrt{\sqrt{2}-1}$

$x^{2}=\sqrt{2}-1$

$(x^{2}+1)^{2}=2$

$x^{4}+2x^{2}-1=0$

## Mathematics Subject Classification

30-00*no label found*12D99

*no label found*

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