Solution - Nonlinear equations

Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "x7"   was replaced by   "x^7". 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((x4) -  (22 • (x2))) -  23x77  = 0 
 Step  2  :

Equation at the end of step  2  :

  ((x4) -  (2•11x2)) -  23x77  = 0 

Step  3  :

Step  4  :

Pulling out like terms :

 4.1     Pull out like factors :

   -8x⁷⁷ + x⁴ - 22x²  =   -x² • (8x⁷⁵ - x² + 22) 

Equation at the end of step  4  :

  -x2 • (8x75 - x2 + 22)  = 0 

Step  5  :

Theory - Roots of a product :

 5.1    A product of several terms equals zero. 

 When a product of two or more terms equals zero, then at least one of the terms must be zero. 

 We shall now solve each term = 0 separately 

 In other words, we are going to solve as many equations as there are terms in the product 

 Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

 5.2      Solve  :    -x² = 0 

 Multiply both sides of the equation by (-1) :  x² = 0


 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      x  =  ± √ 0  

 Any root of zero is zero. This equation has one solution which is  x = 0

Equations of order 5 or higher :

 5.3     Solve   8x⁷⁵-x²+22 = 0

Points regarding equations of degree five or higher.

 (1)  There is no general method (Formula) for solving polynomial equations of degree five or higher.

 (2)  By the Fundamental theorem of Algebra, if we allow complex numbers, an equation of degree  n  will have exactly  n  solutions
(This is if we count double solutions as  2 , triple solutions as  3  and so on

) (3)  By the Abel-Ruffini theorem, the solutions can not always be presented in the conventional way using only a finite amount of additions, subtractions, multiplications, divisions or root extractions

 (4)  If  F(x)  is a polynomial of odd degree with real coefficients, then the equation  F(X)=0  has at least one real solution.

 (5)  Using methods such as the  Bisection  Method, real solutions can be approximated to any desired degree of accuracy. Failed to find the initial interval for implementing the BiSection Method

One solution was found :