Solution - Equations reducible to quadratic form

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((x6) -  (11 • (x4))) +  (2•32x2)  = 0 
 Step  2  :

Equation at the end of step  2  :

  ((x6) -  11x4) +  (2•32x2)  = 0 

Step  3  :

Step  4  :

Pulling out like terms :

 4.1     Pull out like factors :

   x⁶ - 11x⁴ + 18x²  =   x² • (x⁴ - 11x² + 18) 

Trying to factor by splitting the middle term

 4.2     Factoring  x⁴ - 11x² + 18 

The first term is,  x⁴  its coefficient is  1 .
The middle term is,  -11x²  its coefficient is  -11 .
The last term, "the constant", is  +18 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 18 = 18 

Step-2 : Find two factors of  18  whose sum equals the coefficient of the middle term, which is   -11 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -9  and  -2 
                     x⁴ - 9x² - 2x² - 18

Step-4 : Add up the first 2 terms, pulling out like factors :
                    x² • (x²-9)
              Add up the last 2 terms, pulling out common factors :
                    2 • (x²-9)
Step-5 : Add up the four terms of step 4 :
                    (x²-2)  •  (x²-9)
             Which is the desired factorization

Trying to factor as a Difference of Squares :

 4.3      Factoring:  x²-2 

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 : 2 is not a square !!

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

Trying to factor as a Difference of Squares :

 4.4      Factoring:  x²-9 

Check : 9 is the square of 3
Check :  x²  is the square of  x¹ 

Factorization is :       (x + 3)  •  (x - 3) 

Equation at the end of step  4  :

  x2 • (x2 - 2) • (x + 3) • (x - 3)  = 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 

  Solution is  x² = 0

Solving a Single Variable Equation :

 5.3      Solve  :    x²-2 = 0 

 Add  2  to both sides of the equation : 
                      x² = 2
 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      x  =  ± √ 2  

 The equation has two real solutions  
 These solutions are  x = ± √2 = ± 1.4142  
 

Solving a Single Variable Equation :

 5.4      Solve  :    x+3 = 0 

 Subtract  3  from both sides of the equation : 
                      x = -3

Solving a Single Variable Equation :

 5.5      Solve  :    x-3 = 0 

 Add  3  to both sides of the equation : 
                      x = 3

Supplement : Solving Quadratic Equation Directly

Solving    x4-11x2+18  = 0   directly 

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Solving a Single Variable Equation :

Equations which are reducible to quadratic :

 6.1     Solve   x⁴-11x²+18 = 0

This equation is reducible to quadratic. What this means is that using a new variable, we can rewrite this equation as a quadratic equation Using  w , such that  w = x²  transforms the equation into :
 w²-11w+18 = 0

Solving this new equation using the quadratic formula we get two real solutions :
   9.0000  or   2.0000

Now that we know the value(s) of  w , we can calculate  x  since  x  is  √ w  

Doing just this we discover that the solutions of
   x⁴-11x²+18 = 0
  are either : 
  x =√ 9.000 = 3.00000  or :
  x =√ 9.000 = -3.00000  or :
  x =√ 2.000 = 1.41421  or :
  x =√ 2.000 = -1.41421

5 solutions were found :

1.  x = 3
2.  x = -3
3.  x = ± √2 = ± 1.4142
4.  x² = 0