Solution - Equations reducible to quadratic form

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((x6) +  22x3) +  8  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  x⁶+4x³+8 

The first term is,  x⁶  its coefficient is  1 .
The middle term is,  +4x³  its coefficient is  4 .
The last term, "the constant", is  +8 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 8 = 8 

Step-2 : Find two factors of  8  whose sum equals the coefficient of the middle term, which is   4 .

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Equation at the end of step  2  :

  x6 + 4x3 + 8  = 0 

Step  3  :

Solving a Single Variable Equation :

Equations which are reducible to quadratic :

 3.1     Solve   x⁶+4x³+8 = 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 = x³  transforms the equation into :
 w²+4w+8 = 0

Solving this new equation using the quadratic formula we get two imaginary solutions :
   w = -2.0000 ± 2.0000 i 
Now that we know the value(s) of  w , we can calculate  x  since  x  is  ∛ w  

Since we are speaking 3rd root, each of the two imaginary solutions of has 3 roots

Tiger finds these roots using de Moivre's Formula

The 3rd roots of  -2.000 + 2.000 i   are:

  x =  1.000 + 1.000 i 
  x = -1.366 + 0.366 i 
  x =  0.366 -1.366 i 

3rd roots of  -2.000- 2.000 i  :
  x =  0.366  + 1.366 i  
  x = -1.366 - 0.366 i 
  x =  1.000 - 1.000 i 
 

6 solutions were found :

1.   x = 1.000 - 1.000 i
2.   x = -1.366 - 0.366 i
3.   x = 0.366 + 1.366 i
4.   x = 0.366 -1.366 i
5.   x = -1.366 + 0.366 i
6.   x = 1.000 + 1.000 i