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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.

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.


3 Answers

Askiitian.Expert Rajat
24 Points
12 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.\,




Q(0) = 0\,,

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


Evident roots: 1 and −1 and −k




    Q(1) = 0\,,    

so 1 is a root. Similarly, if


that is,


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 \,


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


 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} .


 \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.



Askiitians Expert

Asha Ram Gairola
13 Points
5 years ago
Dear Askiitians Expert, 
Many thanks for this method, I found very interesting. I belive that it could be generelized to higher degree equations too. 
Rishi Sharma
askIITians Faculty 646 Points
one year ago
Dear Student,
Please find below the solution to your problem.


Thanks and Regards

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