Solution - Linear equations with one unknown

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                     x^29-(6*x)=0 

Step by step solution :

Step  1  :

Step  2  :

Pulling out like terms :

 2.1     Pull out like factors :

   x²⁹ - 6x  =   x • (x²⁸ - 6) 

Trying to factor as a Difference of Squares :

 2.2      Factoring:  x²⁸ - 6 

Theory : A difference of two perfect squares,  A² - B²  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =
         A² - AB + BA - B² =
         A² - AB + AB - B² =
         A² - B²

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check : 6 is not a square !!

Ruling : Binomial can not be factored as the difference of two perfect squares.

Equation at the end of step  2  :

  x • (x28 - 6)  = 0 

Step  3  :

Theory - Roots of a product :

 3.1    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 :

 3.2      Solve  :    x = 0 

  Solution is  x = 0

Solving a Single Variable Equation :

 3.3      Solve  :    x²⁸-6 = 0 

 Add  6  to both sides of the equation : 
                      x²⁸ = 6
                     x  =  28th root of (6) 

 The equation has two real solutions  
 These solutions are  x = ± 28th root of 6 = ± 1.0661  
 

Three solutions were found :

1.  x = ± 28th root of 6 = ± 1.0661
2.  x = 0