§ **3 ****Special solutions to algebraic equations**

Abel proved that there are no algebraic solutions to general equations of degree 5 and higher . But Abel's theorem does not answer the question: is there an algebraic solution to each given specific equation . Galois proved that there is an algebraic solution A specific integer-coefficient algebraic equation that cannot be solved by the method . For example

*x *^{5} - *x* +1=0

Galois also found necessary and sufficient conditions for equations to be solved by radicals .

1. Find rational roots

According to the properties of "whole roots and rational roots" in the previous section, the rational roots of some specific equations can be found .

Find the rational roots of an equation . _{}

Solution Since *p* and *q* of the rational roots of this equation are both divisors of 2 , they are 1 , -1 , 2 , and -2 . So the possible values are 1 , -1 , 2 , and -2 . Divide by synthesis ( See § 2 , a ) _{}_{}_{}

test:

1) 2 - 3 2 2 - 1) 2 - 3 2 2

2-1 1-2 5-7 _ _ _ _ _ _

2 - 1 1 3 2 - 5 7 - 5

_{} 2 - 3 2 2 2
_{} - 3
2 2

1 - 1 _{} - 1
2 - 2

2 - 2 1 2 _{} - 4
4 0

So it is a rational root of the known equation ._{}

Divide the original formula by , get the quotient formula_{}

_{} which is _{}

The tower top discriminant 4 - 8<0 , its two roots are a pair of conjugate complex roots . Therefore, the original equation has only one rational root ._{}

2. Solve three equations

Shaped like

*au *^{2 n} + *bu ^{n} +c* =0

The equation of is called a three-term equation, where *a, b, c, n* are not equal to zero , *n* is an integer . It can be solved by radicals . Let *u ^{n}* =

Solve the equation by example

_{}

Solve the order , then get , its root is and . From get . So. Substitute into the original equation to check, we can see that these four numbers are the roots of the equation . _{}_{}_{}_{}_{}_{}_{}

3. Solve the reciprocal equation

Shaped like

*ax ^{n}* +

(where ^{the}* coefficients of the xn ^{-k}* and

Divide both sides of the 1° even-numbered ( *n* = 2 *k* ) reciprocal equation by *x ^{k}* , and then set

*x *^{2} - *zx* +1=0

ask for.

2°
Solution of odd-numbered ( *n* = 2 *k* +1) reciprocal equations boils down to solving even-ordered reciprocal equations .

Solve the equation by example

_{}

The solution is a root of the original equation, and the equation is divided by the 4th reciprocal equation: _{}_{}

_{}

Divide it by , and then combine, we get_{}

_{}

Let , then , the above formula becomes_{}_{}

_{}

From this, there are two definite equations:_{}_{}

_{} and _{}

From this we get

_{}

4. Solve the binomial equation

In the form of *x ** ^{n}* -

The equation of is called a binomial equation . Its *n* roots are the *nth** roots of the complex number **A.*

If *A* is written as

*A* = *r* ( cos *θ* + *i* sin *θ* )

Then the *n* roots of the equation *x ^{n}* -

_{} ( *k* = 0 , 1, 2 , L , *n* - 1)

Geometric Explanation: The points on the complex plane that correspond to the nth root of the number r *(* cos θ + i sin θ ) are the vertices of a regular *n* - *gon* , *and* these *vertices* are *on* a circle with the origin as the center and the radius . And One of the vertices of this *n* -gon has an argument . Figure 3.1 shows the case of *n* = 6 ._{}_{}

If *A* =1 , then the solution *ξ of **x ^{n}* =1 is called

cos + *i* sin ( *k* = 0,1,2, L , *n* - 1) Figure 3.1_{}_{}

If *ξ* is one of the *n* -th unit roots, then the *n **n* - th unit roots are 1, *ξ **, **ξ *^{2 }*, **L** , **ξ *^{n}^{ - }^{1} , which are geometrically represented as vertices of an inscribed regular *n* -gon of the unit circle .* *^{}^{}^{}

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