I wonder how could I solve a biquadratic equations of general structure "ax^4+bx^3+cx^2+dx+e" . Can any one of you pls help me.

(\/)ur@l! Kr!$hn@ k, 14 years ago

Grade:12

3 Answers

Askiitian.Expert Rajat

24 Points

14 years ago

Solving a quartic equation

Special cases

Consider the quartic

$Q(x) = a_4x^4+a_3x^3+a_2x^2+a_1x+a_0.\,$

$a_0=0,\,$

then

$Q(0) = 0\,$ ,

so zero is a root. To find the other roots, we can divide by $x\,$ and solve the resultin cubic equation

$a_4x^3+a_3x^2+a_2x+a_1=0.\,$

Evident roots: 1 and −1 and −k

$a_0+a_1+a_2+a_3+a_4=0,\,$

then

$Q(1) = 0\,$ ,

so 1 is a root. Similarly, if

$a_4-a_3+a_2-a_1+a_0=0,\,$

that is,

$a_0+a_2+a_4=a_1+a_3,\,$

then -1 is a root.

When 1 is a root, we can divide

$Q(x)\,$ by $x-1\,$

and get

$Q(x) = (x - 1)p(x),\,$

where $p(x)\,$ is a cubic polynomial, which may be solved to find $Q\,$ 's other roots. Similarly, if -1 is a root,

$Q(x) = (x + 1)p(x),\,$

where $p(x)\,$ is some cubic polynomial.

$a_2 = 0, a_3 = ka_4, a_0 = ka_1,\,$

then −k is a root and we can factor out $x+k\,$ ,

$\begin{align} Q(x) &=a_4 x^4 + k a_4 x^3 + a_1 x + ka_1 \&=(x + k)a_4x^3 + (x + k)a_1 \&=(x + k)(a_4x^3 + a_1). \end{align}$

And if

$a_0=0, a_3=ka_4, a_1 = ka_2,\,$

then both $0\,$ and $-k\,$ are roots Now we can factor out $x(x + k)\,$ and get

$\begin{align} Q(x) &=x(a_4x^3+ka_4x^2+a_2x+ka_2) \&=x(x+k)(a_4x^2+a_2). \end{align}$

To get Q 's other roots, we simply solve the quadratic factor.

Biquadratic equations

If $a_3=a_1=0,\,$ then

$Q(x) = a_4x^4+a_2x^2+a_0.\,\!$

We call such a polynomial a biquadratic, which is easy to solve.

Let $z=x^2.\,$ Then Q becomes a quadratic q in z,

$q(z) = a_4z^2+a_2z+a_0.\,\!$

Let $z_+\,$ and $z_-\,$ be the roots of q. Then the roots of our quartic Q are

$\begin{align} x_1&=+\sqrt{z_+}, \x_2&=-\sqrt{z_+}, \x_3&=+\sqrt{z_-}, \x_4&=-\sqrt{z_-}. \end{align}$

Quasi-symmetric equations

$a_0x^4+a_1x^3+a_2x^2+a_1 m x+a_0 m^2=0 \,$

Steps:

1) Divide by x².

2) Use variable change z = x + m/x.

The general case, along Ferrari's lines

To begin, the quartic must first be converted to a depressed quartic.

Converting to a depressed quartic

Let

$A x^4 + B x^3 + C x^2 + D x + E = 0 \qquad\qquad(1')$

be the general quartic equation which it is desired to solve. Divide both sides by A,

$x^4 + {B \over A} x^3 + {C \over A} x^2 + {D \over A} x + {E \over A} = 0.$

The first step should be to eliminate the x³ term. To do this, change variables from x to u, such that

$x = u - {B \over 4 A}$ .

Then

$\left( u - {B \over 4 A} \right)^4 + {B \over A} \left( u - {B \over 4 A} \right)^3 + {C \over A} \left( u - {B \over 4 A} \right)^2 + {D \over A} \left( u - {B \over 4 A} \right) + {E \over A} = 0.$

Expanding the powers of the binomials produces

$\left( u^4 - {B \over A} u^3 + {6 u^2 B^2 \over 16 A^2} - {4 u B^3 \over 64 A^3} + {B^4 \over 256 A^4} \right) + {B \over A} \left( u^3 - {3 u^2 B \over 4 A} + {3 u B^2 \over 16 A^2} - {B^3 \over 64 A^3} \right) + {C \over A} \left( u^2 - {u B \over 2 A} + {B^2 \over 16 A^2} \right) + {D \over A} \left( u - {B \over 4 A} \right) + {E \over A} = 0.$

Collecting the same powers of u yields

$u^4 + \left( {-3 B^2 \over 8 A^2} + {C \over A} \right) u^2 + \left( {B^3 \over 8 A^3} - {B C \over 2 A^2} + {D \over A} \right) u + \left( {-3 B^4 \over 256 A^4} + {C B^2 \over 16 A^3} - {B D \over 4 A^2} + {E \over A} \right) = 0.$

Now rename the coefficients of u. Let

$\begin{align} \alpha & = {-3 B^2 \over 8 A^2} + {C \over A} ,\\beta & = {B^3 \over 8 A^3} - {B C \over 2 A^2} + {D \over A} ,\\gamma & = {-3 B^4 \over 256 A^4} + {C B^2 \over 16 A^3} - {B D \over 4 A^2} + {E \over A} . \end{align}$

The resulting equation is

$u^4 + \alpha u^2 + \beta u + \gamma = 0 \qquad \qquad (1)$

which is a depressed quartic equation.

If $\beta=0 \$ then we have a biquadratic Equation, which (as explained above) is easily solved; using reverse substitution we can find our values for $x$ .