Quartic equation

In mathematics, a quartic equation is the result of setting a quartic function equal to zero. An example of a quartic equation is the equation

Solving the quartic equation

According to the fundamental theorem of algebra, a quartic equation always has four solutions (roots). They may be complex and there may be duplicate solutions.

Related Topics:
Fundamental theorem of algebra - Root - Complex

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Much effort has been turned to finding these roots. As with other polynomials, it is sometimes possible to factor a quartic equation directly; but more often such a feat is Herculean, especially when the roots are irrational or complex. Hence it would be useful to have a general formula or algorithm (analogous to the quadratic equation, which solves all quadratics). After much effort, such a formula was indeed found for quartics — but since then it has been proven (by Evariste Galois) that such an approach dead-ends with quartics; they are the highest-degree polynomial equations whose roots can be expressed in a formula using a finite number of arithmetic operators and n-th roots. From quintics on up, one requires more powerful methods if a general algebraic solution is sought, as explained under quintic equations.

Related Topics:
Polynomial - Quadratic equation - Evariste Galois - Quintic

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Given the complexity of the quartic formulae (see below), they are not often used. If only the real rational roots are needed, they can be found (as is true for polynomials of any degree) via trial and error, using Ruffini's rule (so long as all the polynomial coefficients are rational). In the modern age of computers, furthermore, good numerical approximations for the roots are rapidly obtainable via Newton's method. But if the quartic must be solved entirely and precisely, the procedures are outlined below.

Related Topics:
Ruffini's rule - Computers - Newton's method

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Special cases

Quartics in name only

If a4 = 0, then one of the roots is x = 0, and the other roots can be found by dividing by x, and solving the resulting cubic equation,

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:a_0x^3+a_1x^2+a_2x+a_3=0.

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Biquadratic equations

A quartic equation where a3 and a1 are equal to 0 takes the form

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:a_0x^4+a_2x^2+a_4=0,!

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and thus is a biquadratic equation, very easy to solve. Let z=x^2, so our equation turns to

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:a_0z^2+a_2z+a_4=0,!

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which is a simple quadratic equation, whose solutions are easily found using the quadratic formula:

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:z={{-a_2pmsqrt{a_2^2-4a_0a_4}} over {2a_0}},!

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When we've solved it (i.e. found these two z values), we can extract x from them

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:x_1=+sqrt{z_1},!

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:x_2=-sqrt{z_1},!

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:x_3=+sqrt{z_2},!

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:x_4=-sqrt{z_2},!

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If any of the z solutions were negative or complex numbers, some of the x solutions are complex numbers.

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The general case, along Ferrari's lines

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

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Converting to a depressed quartic

Let

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: A x^4 + B x^3 + C x^2 + D x + E = 0 qquadqquad(1')

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

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

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

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: x = u - {B over 4 A} .

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Then

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

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Expanding the powers of the binomials produces

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: 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} ight)

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+ {B over A}

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left( u^3 - {3 u^2 B over 4 A} + {3 u B^2 over 16 A^2} - {B^3 over 64 A^3} ight)

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+ {C over A}

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left( u^2 - {u B over 2 A} + {B^2 over 16 A^2} ight) + {D over A} left( u - {B over 4 A} ight) + {E over A}.

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Collecting the same powers of u yields

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: u^4 + left( {-3 B^2 over 8 A^2} + {C over A} ight) u^2 + left( {B^3 over 8 A^3} - {B C over 2 A^2} + {D over A} ight) 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} ight) = 0.

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Now rename the coefficients of u. Let

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: lpha = {-3 B^2 over 8 A^2} + {C over A},

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: eta = {B^3 over 8 A^3} - {B C over 2 A^2} + {D over A},

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: gamma = {-3 B^4 over 256 A^4} + {C B^2 over 16 A^3} - {B D over 4 A^2} + {E over A}.

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The resulting equation is

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: u^4 + lpha u^2 + eta u + gamma = 0 qquad qquad (1)

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which is a depressed quartic equation.

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If eta=0 then we have a Biquadratic equation, which (as explained above) is easily solved; using reverse substitution we can find our values for x.

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Ferrari's solution

Otherwise, the depressed quartic can be solved by means of a method discovered by Ferrari. Once the depressed quartic has been obtained, the next step is to add the valid identity

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: (u^2 + lpha)^2 - u^4 - 2 lpha u^2 = lpha^2

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to equation (1), yielding

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: (u^2 + lpha)^2 + eta u + gamma = lpha u^2 + lpha^2. qquad qquad (2)

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The effect has been to fold up the u4 term into a perfect square: (u2 + α)2. The second term, α u2 did not disappear, but its sign has changed and it has been moved to the right side.

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The next step is to insert a variable y into the perfect square on the left side of equation (2), and a corresponding 2y into the coefficient of u2 in the right side. To accomplish these insertions, the following valid formulas will be added to equation (2),

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:

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egin{matrix}

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(u^2+lpha+y)^2-(u^2+lpha)^2 & = & 2y(u^2+lpha)+ y^2 \

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\

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& = & 2yu^2+2ylpha+y^2,

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end{matrix}

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and

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: 0 = (lpha + 2 y) u^2 - 2 y u^2 - lpha u^2

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These two formulas, added together, produce

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: (u^2 + lpha + y)^2 - (u^2 + lpha)^2 = (lpha + 2 y) u^2 - lpha u^2 + 2 y lpha + y^2 qquad qquad (y-hbox{insertion})

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which added to equation (2) produces

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: (u^2 + lpha + y)^2 + eta u + gamma = (lpha + 2 y) u^2 + (2 y lpha + y^2 + lpha^2).

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This is equivalent to

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: (u^2 + lpha + y)^2 = (lpha + 2 y) u^2 - eta u + (y^2 + 2 y lpha + lpha^2 - gamma). qquad qquad (3)

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The objective now is to choose a value for y such that the right side of equation (3) becomes a perfect square. This can be done by letting the discriminant of the quadratic function become zero. To explain this, first expand a perfect square so that it equals a quadratic function:

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: (s u + t)^2 = (s^2) u^2 + (2 s t) u + (t^2).

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The quadratic function on the right side has three coefficients. It can be verified that squaring the second coefficient and then subtracting four times the product of the first and third coefficients yields zero:

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: (2 s t)^2 - 4 (s^2) (t^2) = 0.

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Therefore to make the right side of equation (3) into a perfect square, the following equation must be solved:

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: (-eta)^2 - 4 (2 y + lpha) (y^2 + 2 y lpha + lpha^2 - gamma) = 0.

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Multiply the binomial with the polynomial,

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: eta^2 - 4 (2 y^3 + 5 lpha y^2 + (4 lpha^2 - 2 gamma) y + (lpha^3 - lpha gamma)) = 0

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Divide both sides by −4, and move the −β2/4 to the right,

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: 2 y^3

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+ 5 lpha y^2

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+ ( 4 lpha^2 - 2 gamma ) y

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+ left( lpha^3 - lpha gamma - {eta^2 over 4} ight)

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