Quadratic equation

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Figure of a fourth grade polynomial function.

In algebra, a quartic equation or quartic equation with one unknown is an algebraic equation that assumes the so-called canonical form:

Fourth grade equation
${displaystyle ax^{4}+bx^{3}+cx^{2}+dx+e=0;,quad aneq 0}$

where a, b, c, d and e ( ${displaystyle aneq 0}$ ) are numbers that belong to a body, usually to the rational $mathbb{Q}$ and occasionally are real numbers or complexes $mathbb{C}$ .

General case

Sea K a body, where you can extract square and cubic roots and therefore also of fourth order, because it is equivalent to extract square roots twice in a row. In this body, it is possible to factor in everything a, and the following identity is valid: ${displaystyle (a-b)^{4}=a^{4}-4a^{3}b+6a^{2}b^{2}-4ab^{3}+b^{4},}$ .

In an algebraically closed field, it is known that every polynomial of degree 4 has four roots. This is the case of the field of complexes, according to the Fundamental Theorem of Algebra.

Quartic equation in finite field

Solve the equation in the whole {0,1,2,3,4,5,6,7,9,10}

{displaystyle x^{4}-x^{2}-240=0}

a root in the finite set of integer remainders of mod 11, that is F[11] is

{displaystyle x=4}

Through the synthetic division remains ${displaystyle (x+1)(x^{3}-x^{2})-240=0}$

Features

If the independent term has a sign - it has at least a real root.
If the complex number ${displaystyle z=a+bi}$ is the root of a quartum equation, it is also its conjugate ${displaystyle z'=a-bi}$ .
The graph of a polynomial function (equation profile) cuts to the X-axis in 0, 1, 2, 3 or 4 points.

A simple case

This quartic equation

{displaystyle x^{4}+x^{3}+x^{2}+x+1=0,}

which is unitary, as a real-valued polynomial never vanishes.

Therefore its four roots are complex, in pairs of conjugates. Precisely the primitive fifth roots of unity. Structured on the breast and cosine basis ${displaystyle {frac {2pi }{5}}}$ radian and his multiples to the room.

Solving methods

There are resolving methods to solve quadratic equations, with which we can reach their solutions, so that the set of real numbers is not algebraically closed, always resulting in four solutions, commonly two real solutions and two complex conjugate solutions (but this may not always be the case). The solutions of the equation can be approximated with the Newton-Raphson method, but only one of the real solutions will be obtained, making this method very disadvantageous due to its limitations in the context of calculus.

Factorization

Sea ${displaystyle P(x)=ax^{4}+bx^{3}+cx^{2}+dx+e}$ the polynomial that wants to find its roots whose coefficients are integer, consider a linear factor ${displaystyle Q(x)=qx-p}$ as one of the dividers of such polynomial, where it is possible to find a quotient ${displaystyle R(x)}$ of third degree that can be resolved by applying factorization again, or by solving it by the method of Cardano (if such cubic quotient is irreducible by rational factors). In making the division ${displaystyle P(x)}$ and $Q(x)$ , we get the quotient ${displaystyle R(x)}$ given

{displaystyle R(x)={frac {a}{q}}x^{3}+{frac {bq+ap}{q^{2}}}x^{2}+{frac {cq^{2}+bpq+ap^{2}}{q^{3}}}x+{frac {dq^{3}+cpq^{2}+bp^{2}q+ap^{3}}{q^{4}}}}

whose resulting residue is:

{displaystyle S(x)={frac {eq^{4}+dpq^{3}+cp^{2}q^{2}+dp^{3}q+ap^{4}}{q^{4}}}}

so if ${displaystyle S(x)=0}$ , then ${displaystyle x_{1}={frac {p}{q}}}$ is a rational root ${displaystyle P(x)}$ and therefore, it is an exact division. However, if ${displaystyle S(x)neq 0}$ , then ${displaystyle P(x)}$ it is an irreducible polynomial, and it must be solved by alternative methods.

Ferrari method

Let be the quartic equation

{displaystyle ax^{4}+bx^{3}+cx^{2}+dx+e=0}

It reduces to the monic form by dividing $a$ :

{displaystyle x^{4}+Bx^{3}+Cx^{2}+Dx+E=0}

where

{displaystyle B={frac {b}{a}},C={frac {c}{a}},D={frac {d}{a}},E={frac {e}{a}}}

Its solving cubic equation is:

{displaystyle y^{3}-Cy^{2}+(BD-4E)y+(4CE-B^{2}E-D^{2})=0}

which can be solved by the method of Cardano, where $y$ is considered a real root of this (no matter if it is a positive or negative real root), the first root being of first priority. However, the nature of the roots of the solving cubic equation will determine the solutions of the original equation, considering the following possibilities:

(1) If the solving cubic equation has a real root, the quartum equation will have two real solutions and two complex conjugated solutions.
(2) If the solving cubic equation has two or three real roots, the quartum equation will have four randomly defined solutions:
- (a) Four different real solutions.
- (b) Two pairs of complex conjugated solutions.
- (c) Two real double roots.
- (d) A simple real root and a triple real root.
- (e) A real quadruple root.
- (f) A real double root and two complex conjugated solutions.
- (g) Two complex double conjugated roots.
- (h) A real double root and two simple real roots.

Once we obtain the positive root of the resolving cubic equation, we calculate the following values:

{displaystyle m={sqrt {{frac {B^{2}}{4}}-C+y}}}

{displaystyle n={sqrt {{frac {y^{2}}{4}}-E}};;qquad Siqquad m=0}

{displaystyle n={frac {By-2D}{4m}};;qquad Siqquad mneq 0}

Of these values, we will solve two quadratic equations:

{displaystyle x^{2}+left({frac {B}{2}}+mright)x+left({frac {y}{2}}-nright)=0}

{displaystyle x^{2}+left({frac {B}{2}}-mright)x+left({frac {y}{2}}+nright)=0}

By solving them by the quadratic formula, we get the solutions of the original quartic equation:

{displaystyle {begin{cases}x_{1}={frac {1}{2}}left[left(-{frac {B}{2}}-mright)+{sqrt {{frac {B^{2}}{4}}+Bm+m^{2}-2y-4n}}right]\x_{2}={frac {1}{2}}left[left(-{frac {B}{2}}-mright)-{sqrt {{frac {B^{2}}{4}}+Bm+m^{2}-2y-4n}}right]\x_{3}={frac {1}{2}}left[left(-{frac {B}{2}}+mright)+{sqrt {{frac {B^{2}}{4}}-Bm+m^{2}-2y+4n}}right]\x_{4}={frac {1}{2}}left[left(-{frac {B}{2}}+mright)-{sqrt {{frac {B^{2}}{4}}-Bm+m^{2}-2y+4n}}right]end{cases}}}

Descartes Method

Let be the quartic equation

{displaystyle ax^{4}+bx^{3}+cx^{2}+dx+e=0}

We divide the initial equation by the quartic component, we obtain:

{displaystyle x^{4}+{frac {b}{a}}x^{3}+{frac {c}{a}}x^{2}+{frac {d}{a}}x+{frac {e}{a}}=0,}

We proceed to perform a transformation of Tschirnhaus, that is, to replace ${displaystyle x=w-{frac {b}{4a}},}$ to convert it into its reduced form:

{displaystyle w^{4}+jw^{2}+kw+l=0,}

whose components are given by:

{displaystyle j={frac {c}{a}}-{frac {3b^{2}}{8a^{2}}}={frac {8ac-3b^{2}}{8a^{2}}}}

{displaystyle k={frac {d}{a}}-{frac {bc}{2a^{2}}}+{frac {b^{3}}{8a^{3}}}={frac {b^{3}-4abc+8a^{2}d}{8a^{3}}}}

{displaystyle l={frac {e}{a}}-{frac {bd}{4a^{2}}}+{frac {b^{2}c}{16a^{3}}}-{frac {3b^{4}}{256a^{4}}}={frac {256a^{3}e-64a^{2}bd+16ab^{2}c-3b^{4}}{256a^{4}}}}

The solving cubic equation of Descartes' method is:

{displaystyle y^{3}+2jy^{2}+(j^{2}-4l)y-k^{2}=0,}

However, unlike the solving cubic equation of the Ferrari method, one of its real roots must be positive, with which we will solve two quadratic equations:

{displaystyle w^{2}+{sqrt {y}}w+{frac {j+y-{frac {k}{sqrt {y}}}}{2}}=0}

{displaystyle w^{2}-{sqrt {y}}w+{frac {j+y+{frac {k}{sqrt {y}}}}{2}}=0}

By solving them by the quadratic formula, then the solutions of the reduced quartic equation are (ordering them by positive and negative signs):

{displaystyle {begin{cases}w_{1}={frac {1}{2}}left({sqrt {y}}+{sqrt {-y-2j-{frac {2k}{sqrt {y}}}}}right)\w_{2}={frac {1}{2}}left({sqrt {y}}-{sqrt {-y-2j-{frac {2k}{sqrt {y}}}}}right)\w_{3}={frac {1}{2}}left(-{sqrt {y}}+{sqrt {-y-2j+{frac {2k}{sqrt {y}}}}}right)\w_{4}={frac {1}{2}}left(-{sqrt {y}}-{sqrt {-y-2j+{frac {2k}{sqrt {y}}}}}right)end{cases}}}

Since the objective is to find the solutions of the original equation, we use the following formula:

{displaystyle x_{n}=w_{n}-{frac {b}{4a}}qquad mathrm {donde} qquad n=1,2,3,4.}

Therefore, we replace $w$ in the formula for ${displaystyle n=1,2,3,4}$ :

{displaystyle {begin{cases}x_{1}={frac {1}{2}}left({sqrt {y}}+{sqrt {-y-2j-{frac {2k}{sqrt {y}}}}}right)-{frac {b}{4a}}\x_{2}={frac {1}{2}}left({sqrt {y}}-{sqrt {-y-2j-{frac {2k}{sqrt {y}}}}}right)-{frac {b}{4a}}\x_{3}={frac {1}{2}}left(-{sqrt {y}}+{sqrt {-y-2j+{frac {2k}{sqrt {y}}}}}right)-{frac {b}{4a}}\x_{4}={frac {1}{2}}left(-{sqrt {y}}-{sqrt {-y-2j+{frac {2k}{sqrt {y}}}}}right)-{frac {b}{4a}}end{cases}}}

Cardano-Vieta Relations

On the other hand, if we use the Cardano-Vieta relations in the solutions of the original quartic equation, we can have the cubic, quadratic, linear components and the independent term in the original equation:

{displaystyle {begin{cases}x_{1}+x_{2}+x_{3}+x_{4}=-{frac {b}{a}}\x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4}={frac {c}{a}}\x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4}=-{frac {d}{a}}\x_{1}x_{2}x_{3}x_{4}={frac {e}{a}}end{cases}}}

Special cases

Bisquare Equations

These are a particular case of the above, whose polynomial form is:

{displaystyle ax^{4}+cx^{2}+e=0}

To solve these equations you just have to make the variable change ${displaystyle x^{2}=t}$ With what we have left:

{displaystyle at^{2}+ct+e=0}

The result turns out to be a quadratic equation that we can solve using the quadratic formula:

{displaystyle t={frac {-cpm {sqrt {c^{2}-4ae}}}{2a}}}

The change of variable is undone to obtain the four solutions:

{displaystyle x_{1}=+{sqrt {t_{1}}}}

{displaystyle x_{2}=-{sqrt {t_{1}}}}

{displaystyle x_{3}=+{sqrt {t_{2}}}}

{displaystyle x_{4}=-{sqrt {t_{2}}}}

Obtaining an equation from a root

Sea $x_{0}$ a root whose value is known:

${displaystyle x={sqrt {2+{sqrt {2}}}}}$
Undoing roots with powers: ${displaystyle x^{2}=2+{sqrt {2}}}$ ${displaystyle x^{2}-2={sqrt {2}}}$ ${displaystyle (x^{2}-2)^{2}=2}$ ${displaystyle x^{4}-4x^{2}+2=0}$ The other roots are: ${displaystyle x_{2}=-{sqrt {2+{sqrt {2}}}}}$ , ${displaystyle x_{3}={sqrt {2-{sqrt {2}}}}}$ and ${displaystyle x_{4}=-{sqrt {2-{sqrt {2}}}}}$ .

Quasisymmetric Equations

The following type of equation

{displaystyle x^{4}+a_{1}x^{3}+a_{2}x^{2}+a_{3}x+m^{2}=0,}

Where

{displaystyle m={frac {a_{3}}{a_{1}}},}

can be resolved like this:

By dividing the equation by $x^{2}$ , you get

{displaystyle x^{2}+{frac {m^{2}}{x^{2}}}+a_{1}x+{frac {a_{3}}{x}}+a_{2}=0}

{displaystyle (x^{2}+{frac {m^{2}}{x^{2}}})+a_{1}(x+{frac {m}{x}})+a_{2}=0}

Making change of variable:

{displaystyle z=x+{frac {m}{x}}}

we got to

{displaystyle z^{2}-2m=x^{2}+{frac {m^{2}}{x^{2}}},}

like this:

{displaystyle (z^{2}-2m)+a_{1}z+a_{2}=0,}

This equation gives 2 roots, ${displaystyle z_{1}}$ and ${displaystyle z_{2}}$ .

The roots of the original equation can be obtained by solving the following quadratic equations:

{displaystyle {begin{cases}x^{2}-z_{1}x+m=0\x^{2}-z_{2}x+m=0end{cases}}}

Yeah. $a_{0}$ is not equal to one in ${displaystyle a_{0}x^{4}+a_{1}x^{3}+a_{2}x^{2}+a_{3}x+a_{0}m^{2}=0,}$ , this method is applicable anyway, after dividing the equation between $a_{0}$ .

Quasisimetric equations possess the following property, which, on the other hand, defines them: $x_{1}$ , $x_{2}$ and ${displaystyle x_{3}}$ , $x_{4}$ are the roots of the equation, then ${displaystyle x_{1}x_{2}=m}$ . Since the product of the 4 roots is $m^{2}$ , then ${displaystyle x_{3}x_{4}=m}$ necessarily.

Symmetric quadratic equations

They have shape ${displaystyle ax^{4}+bx^{3}+cx^{2}+bx+a=0}$ with a 0. All coefficients are rational numbers.

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