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(\/)ur@l! Kr!$hn@ k Grade: 12
        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.
7 years ago

Answers : (2)

Askiitian.Expert Rajat
24 Points
										

Solving a quartic equation


Special cases


Consider the quartic


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

If


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


If  


   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.


If


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 x 2.


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.


If \gamma=0 \ then one of the roots is u=0 \ and the other roots can be found by dividing by u, and solving the resulting equation,


 u^3 + \alpha u + \beta = 0 \,.

Using reverse substitution we can find our values for x.


Regards,


Rajat,


Askiitians Expert

7 years ago
Asha Ram Gairola
13 Points
										
Dear Askiitians Expert, 
Many thanks for this method, I found very interesting. I belive that it could be generelized to higher degree equations too. 
7 months ago
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