Solution - Equations reducible to quadratic form

Step by step solution :

Step  1  :

Equation at the end of step  1  :

                    2  
  ((x2) -  3) +  ——————  = 0 
                 (x2^2)
 Step  2  :
             2
 Simplify   ——
            x4

Equation at the end of step  2  :

                  2
  ((x2) -  3) +  ——  = 0 
                 x4

Step  3  :

Rewriting the whole as an Equivalent Fraction :

 3.1   Adding a fraction to a whole

Rewrite the whole as a fraction using  x⁴  as the denominator :

               x2 - 3     (x2 - 3) • x4
     x2 - 3 =  ——————  =  —————————————
                 1             x4      

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Trying to factor as a Difference of Squares :

 3.2      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.

Adding fractions that have a common denominator :

 3.3       Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

 (x2-3) • x4 + 2     x6 - 3x4 + 2
 ———————————————  =  ————————————
       x4                 x4     

Polynomial Roots Calculator :

 3.4    Find roots (zeroes) of :       F(x) = x⁶ - 3x⁴ + 2
Polynomial Roots Calculator is a set of methods aimed at finding values of  x  for which   F(x)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  x  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

In this case, the Leading Coefficient is  1  and the Trailing Constant is  2.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1 ,2

 Let us test ....

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms

In our case this means that
   x⁶ - 3x⁴ + 2 
can be divided by 2 different polynomials,including by  x - 1 

Polynomial Long Division :

 3.5    Polynomial Long Division
Dividing :  x⁶ - 3x⁴ + 2 
                              ("Dividend")
By         :    x - 1    ("Divisor")

Quotient :  x⁵+x⁴-2x³-2x²-2x-2  Remainder:  0 

Polynomial Roots Calculator :

 3.6    Find roots (zeroes) of :       F(x) = x⁵+x⁴-2x³-2x²-2x-2

     See theory in step 3.4
In this case, the Leading Coefficient is  1  and the Trailing Constant is  -2.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1 ,2

 Let us test ....

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms

In our case this means that
   x⁵+x⁴-2x³-2x²-2x-2 
can be divided with  x+1 

Polynomial Long Division :

 3.7    Polynomial Long Division
Dividing :  x⁵+x⁴-2x³-2x²-2x-2 
                              ("Dividend")
By         :    x+1    ("Divisor")

Quotient :  x⁴-2x²-2  Remainder:  0 

Trying to factor by splitting the middle term

 3.8     Factoring  x⁴-2x²-2 

The first term is,  x⁴  its coefficient is  1 .
The middle term is,  -2x²  its coefficient is  -2 .
The last term, "the constant", is  -2 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -2 = -2 

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

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

Equation at the end of step  3  :

  (x4 - 2x2 - 2) • (x + 1) • (x - 1)
  ——————————————————————————————————  = 0 
                  x4                

Step  4  :

When a fraction equals zero :

 4.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

  (x4-2x2-2)•(x+1)•(x-1)
  —————————————————————— • x4 = 0 • x4
            x4          

Now, on the left hand side, the  x⁴  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   (x⁴-2x²-2)  •  (x+1)  •  (x-1)  = 0

Theory - Roots of a product :

 4.2    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 :

Equations which are reducible to quadratic :

 4.3     Solve   x⁴-2x²-2 = 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 :
 w²-2w-2 = 0

Solving this new equation using the quadratic formula we get two real solutions :
   2.7321  or  -0.7321

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
   x⁴-2x²-2 = 0
  are either : 
   x =√ 2.732 = 1.65289  or :
   x =√ 2.732 = -1.65289  or :
   x =√-0.732 = 0.0 + 0.85560 i  or :
   x =√-0.732 = 0.0 - 0.85560 i

Solving a Single Variable Equation :

 4.4      Solve  :    x+1 = 0 

 Subtract  1  from both sides of the equation : 
                      x = -1

Solving a Single Variable Equation :

 4.5      Solve  :    x-1 = 0 

 Add  1  to both sides of the equation : 
                      x = 1

6 solutions were found :

1.  x = 1
2.  x = -1
3.  x =√-0.732 = 0.0 - 0.85560 i
4.  x =√-0.732 = 0.0 + 0.85560 i
5.  x =√ 2.732 = -1.65289
6.  x =√ 2.732 = 1.65289