##### 1. CRIMINAL RATIONAL EQUATIONS WHICH ARE REDUCED TO SQUARE EQUATIONS

An equation in which the variable is contained in the denominator of a fraction is called a fractional rational equation.

When solving fractional rational equations, most often, the procedure is:

– the denominators of multipliers are decomposed;

– the definition area of the equation is determined;

– we get rid of the fractions by multiplying the equation by NHS for the denominators;

– the obtained equation is solved;

– it is determined which of the obtained solutions are solutions of the initial equation.

**Example 2. ** Let's solve the equation

The defining area of the equation is the set

**D = R {-2, 2}***.*

Multiply the equation b** y (x – 2) (x + 2**), so we get the equation:

**х (х + 2) – З (х – 2) = **8,

that is, the equation:

*х² – х – 2 = 0*

whose solutions a* re х₁ = -1 *a

**2. Since 2**

*nd х₂ =***it follows that the equation has only one root x**

*∉ D,**1.*

**= –**##### 2. BINOMINAL EQUATIONS

The equation of type

**xⁿ – а = 0**

is calle*d the nth degree *bin*om*ial equation.

With the change it comes down to a simpler equation:

* уⁿ – 1 = 0*

Depending on the coefficient а and the parity of the number n, *f*or the binomial equation х ра* – а = 0, th*e following applies:

1) If** a = **0, the equation in the set of real numbers, ie the set of complex numbers has a unique solution

*x = 0*2) If ** a ≠ 0**, then in the set of real numbers the equation for:

a)** n = 2**k a

**0 has two real roots**

*nd a>*b)** n = 2k** a

**0 has no real roots;**

*nd a <*c** ) n = 2k + 1** a

*n*R) has a single real root

**d (∀a**∈3) If** a ≠ 0** is an arbitrary real number (or complex) number, then there are exactly n roots in the set of complex

*numbers.*

##### 3. BIQUARD EQUATIONS

The equation of type

*ах⁴ + bx² + c = 0*

where a,* b, c ar*e real numbers and a ≠

**called a**

*0 is***biquadratic equatio**n.

We solve the biquadratic equation with the help of the change ** χ² = у** which reduces it to the quadratic equation:

**ay² + by + c = 0**

If the solutions of the equat* ion ay² + by + c = 0 ar*e у₁

**nd у₂, then based on the chan**

*a***у we get the set of equations:**

*ge х² =***x² = y**₁,** x² = y₂**

For *у₁, у₂ ∈ R, *the solutions of the biquadratic equation are:

The biquadratic equation

** аx⁴ + bx² + c =** 0, a

*t c =*0 gets the type:

** ax⁴ + bx² = 0** ie

*x² (ax² + b) = 0*whose solution comes down to solving two incomplete quadratic equations:

** x² = 0 **a

**nd ах² + b =**0.If **b = 0,** then the biquadratic equation

**0 is reduced to a fourth degree binomial equatio**

*аx⁴ + bx² + c =***n**0.

*ах⁴ + с =*For this equation, *at ace*> 0 all four roots are complex, and f*or ace* <0 it has two real and two complex roots.

If b* = c =* 0, then the biquadratic equation has a quadratic root* x = 0.*

##### 4. TRINOMAL EQUATIONS

The equation of type

** ах²ⁿ + bxⁿ + c = 0 **where a, b,

*c ∈ R and*а

*≠ 0 is*called a

**trinomial equatio**n.

The equation is *of* the 2nd degree and its solution with the shift

*хⁿ = у*

comes down to solving the quadratic equation **ay² + by + c = 0**.

If u*₁* and* u*₂ are the roots of the last equation, then the roots of the trinomial equation are obtained by solving the set of binomial equations:

*хⁿ = У₁*

* хⁿ = У₂*

##### 5. SQUARE EQUATIONS

These equations usually have the type:

*a[ f (x)²] + b [ f (x)]+ c = 0*

where *a, b, c ∈ R (a ≠ O),* and *f (x)* is an expression that depends o*n *x.

Such equations are called **quadratic equations, bec**ause the change f (x*) = u co*mes down to a quadratic equation:

**ay² + by + c = 0**

One root of the third degree symmetric equation is always the number -1, and the other two are obtained by solving the quadratic equation:

**ах² + (b – a) x + a = 0**

The third degree equation of type

**ах³ – bx² – bx – a = 0**, (а ≠ 0)

is solved in a similar way as the third degree symmetric equation. The number 1 is always the solution to this equation.

###### 6. FOURTH DEGREE SYMMETRIC EQUATION

The equation of type:

* ах⁴ + bx³ + сх² + bx + а = о, (а ≠ 0),* is called the fourth degree symmetric equation.

By grouping the terms with equal coefficients, we get the equation:

**а (х⁴ + 1) + b (x³ + х) + сх² = 0**

Divide this equation by* x² (x ≠ *0) and get the equation

(*)

We introduce a new variable with the shift, from which squaring is obtained:

that is

By substituting the corresponding values in (*), we get the pawn

*а (у² – 2) + by + c = 0*

If *у₁, у₂ ∈ R* are the roots of the last equation, then the values of x a*r*e found from the equations:

that is, from the set of equations:

Hence the four roots of the symmetric equation are obtained, provided that *у₁, у₂ ∈ R*

The quadratic equation of the type:

**ax⁴ + bx³ ± bx – a = 0 (a ≠ 0**),

is solved by decomposing its left-hand side into factors, giving:

**(х² ± 1) (ах² + bx + a) = 0**

###### 7. FOURTH DEGREE SYMMETRIC EQUATIONS

The equation of form:

* ах⁵ + bx⁴ + сх³ + cx² + bx + а = 0, (a. *0) is a fifth degree symmetric equation.

The solution procedure is as follows:

– the members with the same coefficients are grouped:

**a (x³ + 1) + bx (x³ + 1) + cx² (x + 1) = 0**

* * x³ + 1 = (x + 1) (x² – x + 1) a*nd

*** x⁵ + 1 = (x + 1) (x⁴ – x³ + x² – x + 1)**

the equation takes the form:

*(x + 1)[ ax⁴ + (b-a)x³ + (a – b + c)x² + (b-a) x + a] = 0*

One root is* x₁ = -1,* and the other 4 roots are obtained by solving the fourth degree symmetric equation:

*ax⁴ + (b-a) x³ + (a – b + c) x² + (b-a) x + a = 0*

The fifth degree equation of the type:

* ах⁵ + bx⁴ + сх³ + cx² – bx – а = 0, (a ≠ 0*),

is solved in a similar way as the fifth degree symmetric equation. One solution to this equation is* x = *1. This equation is called the fifth degree obliquely symmetric equation.

– equations that are squared –