Solution - Equations reducible to quadratic form

Rearrange:

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

                     (x^2-6)^2-(-3*x^2+18)=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (((x2)-6)2)-((0-3x2)+18)  = 0 

Step  2  :

 2.1     Evaluate :  (x2-6)2   =    x4-12x2+36 

Trying to factor by splitting the middle term

 2.2     Factoring  x⁴-9x²+18 

The first term is,  x⁴  its coefficient is  1 .
The middle term is,  -9x²  its coefficient is  -9 .
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   -9 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -6  and  -3 
                     x⁴ - 6x² - 3x² - 18

Step-4 : Add up the first 2 terms, pulling out like factors :
                    x² • (x²-6)
              Add up the last 2 terms, pulling out common factors :
                    3 • (x²-6)
Step-5 : Add up the four terms of step 4 :
                    (x²-3)  •  (x²-6)
             Which is the desired factorization

Trying to factor as a Difference of Squares :

 2.3      Factoring:  x²-3 

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 : 3 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 :

 2.4      Factoring:  x²-6 

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  :

  (x2 - 3) • (x2 - 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  :    x²-3 = 0 

 Add  3  to both sides of the equation : 
                      x² = 3
 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      x  =  ± √ 3  

 The equation has two real solutions  
 These solutions are  x = ± √3 = ± 1.7321  
 

Solving a Single Variable Equation :

 3.3      Solve  :    x²-6 = 0 

 Add  6  to both sides of the equation : 
                      x² = 6
 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      x  =  ± √ 6  

 The equation has two real solutions  
 These solutions are  x = ± √6 = ± 2.4495  
 

Supplement : Solving Quadratic Equation Directly

Solving    x4-9x2+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 :

 4.1     Solve   x⁴-9x²+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 :
 w²-9w+18 = 0

Solving this new equation using the quadratic formula we get two real solutions :
   6.0000  or   3.0000

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
   x⁴-9x²+18 = 0
  are either : 
  x =√ 6.000 = 2.44949  or :
  x =√ 6.000 = -2.44949  or :
  x =√ 3.000 = 1.73205  or :
  x =√ 3.000 = -1.73205

Four solutions were found :

1.  x = ± √6 = ± 2.4495
2.  x = ± √3 = ± 1.7321