Biquadratic equations refers to the 4th degree equations that has only quartic, quadratic and constant terms. It is a special pattern of quartic equations that is very similar to quadratic equation. Below is the general form of a biquadratic equation.

y^4+py^2+r=0
(1)

Taking y^2 as a variable, the equation (1) becomes a quadratic equation in terms y^2 . Apply the root formula for a quadratic equation.

y^2 = \dfrac{-p\pm\sqrt{p^2-4r}}{2}
(2)

Then,

y^2 = \dfrac{-p+\sqrt{p^2-4r}}{2}
(3)

or

y^2 = \dfrac{-p-\sqrt{p^2-4r}}{2}
(4)

Solving the equation (3) gives two roots of the biquadratic equation.

y_1 = \sqrt{\dfrac{-p+\sqrt{p^2-4r}}{2}}
(5)
y_2 = -\sqrt{\dfrac{-p+\sqrt{p^2-4r}}{2}}
(6)

Solving the equation (4) gives another two roots of the biquadratic equation.

y_3 = \sqrt{\dfrac{-p-\sqrt{p^2-4r}}{2}}
(7)
y_4 = -\sqrt{\dfrac{-p-\sqrt{p^2-4r}}{2}}
(8)

If p^2>4r, the radicant under the first radical sign is real number. It could be negative, which results in complex root finally. But it will be not a problem.

If p^2< 4r, the root will be the square root of a complex number. Then it will involve the square root of a complex number.

y_1 = \sqrt{\dfrac{-p+i\sqrt{4r-p^2}}{2}}
(9)

Let y_1 = m+ni , that is

m+ni = \sqrt{\dfrac{-p+i\sqrt{4r-p^2}}{2}}
(10)

Square the both sides

m^2-n^2+2mni = - \dfrac{p}{2} +\dfrac{i\sqrt{4r-p^2}}{2}

Comparing the real and imaginary parts gives the following equations

m^2-n^2 = - \dfrac{p}{2}

2mn = \dfrac{\sqrt{4r-p^2}}{2}

This is equivalent to solving the system of equations.

\begin{cases} m^2-n^2 = u \\ 2mn = v \end{cases}
(11)

if u = - \dfrac{p}{2} an v = \dfrac{\sqrt{4r-p^2}}{2}

To solve for m and n, substitute (11) into the following identity

(m^2+n^2)^2 = (m^2-n^2)^2+4m^2n^2

Then,

m^2+n^2 = \sqrt{u^2+v^2}
(12)

Solve the system of equations

\begin{cases} m^2-n^2 = u \\ m^2+n^2 = \sqrt{u^2+v^2} \end{cases}

Then,

m^2 = \dfrac{ \sqrt{u^2+v^2} +u}{2}
(13)
n^2 = \dfrac{ \sqrt{u^2+v^2} -u}{2}
(14)

Substitute the value of u and v

u^2+v^2 = ( - \dfrac{p}{2})^2+(\dfrac{\sqrt{4r-p^2}}{2})^2=r

m^2 = \dfrac{\sqrt{r} - \dfrac{p}{2} }{2} =\dfrac{\sqrt{r}}{2}-\dfrac{p}{4}

m =\pm \sqrt{ \dfrac{\sqrt{r}}{2}-\dfrac{p}{4}}
(15)
n = \pm \sqrt{\dfrac{ \sqrt{r}}{2}+\dfrac{p}{4}}
(16)

y_1=m+ni = \pm( \sqrt{ \dfrac{\sqrt{r}}{2}-\dfrac{p}{4}}+i \sqrt{\dfrac{ \sqrt{r}}{2}+\dfrac{p}{4}})

Similarly, taking square roots for y_2, y_3 and y_4 if p^2< 4r gives the same result.

Now selecting different signs for m and n represents 4 roots. Therefore,

y_1= \sqrt{ \dfrac{\sqrt{r}}{2}-\dfrac{p}{4}}+i \sqrt{\dfrac{ \sqrt{r}}{2}+\dfrac{p}{4}}

y_2= -\sqrt{ \dfrac{\sqrt{r}}{2}-\dfrac{p}{4}}+i \sqrt{\dfrac{ \sqrt{r}}{2}+\dfrac{p}{4}}

y_3= -\sqrt{ \dfrac{\sqrt{r}}{2}-\dfrac{p}{4}}-i \sqrt{\dfrac{ \sqrt{r}}{2}+\dfrac{p}{4}}

y_4= \sqrt{ \dfrac{\sqrt{r}}{2}-\dfrac{p}{4}}-i \sqrt{\dfrac{ \sqrt{r}}{2}+\dfrac{p}{4}}

when p^2< 4r.

Collected in the board: Quartic Equations

Steven Zheng posted 1 month ago

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