Solution - Adding, subtracting and finding the least common multiple

Rearrange:

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

                x/(x-2)+1*(x+2)-(8/(x^2-4))=0 

Step by step solution :

Step  1  :

               8  
 Simplify   ——————
            x2 - 4

Trying to factor as a Difference of Squares :

 1.1      Factoring:  x² - 4 

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 : 4 is the square of 2
Check :  x²  is the square of  x¹ 

Factorization is :       (x + 2)  •  (x - 2) 

Equation at the end of step  1  :

     x                   8     
  (—————+(1•(x+2)))-———————————  = 0 
   (x-2)            (x+2)•(x-2)

Step  2  :

Equation at the end of step  2  :

     x               8     
  (—————+(x+2))-———————————  = 0 
   (x-2)        (x+2)•(x-2)

Step  3  :

              x  
 Simplify   —————
            x - 2

Equation at the end of step  3  :

     x                          8        
  (————— +  (x + 2)) -  —————————————————  = 0 
   x - 2                (x + 2) • (x - 2)

Step  4  :

Rewriting the whole as an Equivalent Fraction :

 4.1   Adding a whole to a fraction

Rewrite the whole as a fraction using  (x-2)  as the denominator :

             x + 2     (x + 2) • (x - 2)
    x + 2 =  —————  =  —————————————————
               1            (x - 2)     

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

Adding fractions that have a common denominator :

 4.2       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:

 x + (x+2) • (x-2)      x2 + x - 4
 —————————————————  =  ———————————
     1 • (x-2)         1 • (x - 2)

Equation at the end of step  4  :

  (x2 + x - 4)            8        
  ———————————— -  —————————————————  = 0 
  1 • (x - 2)     (x + 2) • (x - 2)

Step  5  :

Trying to factor by splitting the middle term

 5.1     Factoring  x²+x-4 

The first term is,  x²  its coefficient is  1 .
The middle term is,  +x  its coefficient is  1 .
The last term, "the constant", is  -4 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -4 = -4 

Step-2 : Find two factors of  -4  whose sum equals the coefficient of the middle term, which is   1 .

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

Calculating the Least Common Multiple :

 5.2    Find the Least Common Multiple

      The left denominator is :       x-2 

      The right denominator is :       (x+2) • (x-2) 

      Least Common Multiple:
      (x-2) • (x+2) 

Calculating Multipliers :

 5.3    Calculate multipliers for the two fractions


    Denote the Least Common Multiple by  L.C.M 
    Denote the Left Multiplier by  Left_M 
    Denote the Right Multiplier by  Right_M 
    Denote the Left Deniminator by  L_Deno 
    Denote the Right Multiplier by  R_Deno 

   Left_M = L.C.M / L_Deno = x+2

   Right_M = L.C.M / R_Deno = 1

Making Equivalent Fractions :

 5.4      Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example :  1/2   and  2/4  are equivalent,  y/(y+1)²   and  (y²+y)/(y+1)³  are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

   L. Mult. • L. Num.      (x2+x-4) • (x+2)
   ——————————————————  =   ————————————————
         L.C.M              (x-2) • (x+2)  

   R. Mult. • R. Num.            8      
   ——————————————————  =   —————————————
         L.C.M             (x-2) • (x+2)

Adding fractions that have a common denominator :

 5.5       Adding up the two equivalent fractions

 (x2+x-4) • (x+2) - (8)     x3 + 3x2 - 2x - 16
 ——————————————————————  =  ——————————————————
     (x-2) • (x+2)          (x - 2) • (x + 2) 

Checking for a perfect cube :

 5.6    x³ + 3x² - 2x - 16  is not a perfect cube

Trying to factor by pulling out :

 5.7      Factoring:  x³ + 3x² - 2x - 16 

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1:  -2x - 16 
Group 2:  x³ + 3x² 

Pull out from each group separately :

Group 1:   (x + 8) • (-2)
Group 2:   (x + 3) • (x²)

Bad news !! Factoring by pulling out fails :

The groups have no common factor and can not be added up to form a multiplication.

Polynomial Roots Calculator :

 5.8    Find roots (zeroes) of :       F(x) = x³ + 3x² - 2x - 16
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  -16.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1 ,2 ,4 ,8 ,16

 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
   x³ + 3x² - 2x - 16 
can be divided with  x - 2 

Polynomial Long Division :

 5.9    Polynomial Long Division
Dividing :  x³ + 3x² - 2x - 16 
                              ("Dividend")
By         :    x - 2    ("Divisor")

Quotient :  x²+5x+8  Remainder:  0 

Trying to factor by splitting the middle term

 5.10     Factoring  x²+5x+8 

The first term is,  x²  its coefficient is  1 .
The middle term is,  +5x  its coefficient is  5 .
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   5 .

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

Canceling Out :

 5.11    Cancel out  (x-2)  which appears on both sides of the fraction line.

Equation at the end of step  5  :

  x2 + 5x + 8
  ———————————  = 0 
     x + 2   

Step  6  :

When a fraction equals zero :

 6.1    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:

  x2+5x+8
  ——————— • x+2 = 0 • x+2
   (x+2) 

Now, on the left hand side, the  x+2  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   x²+5x+8  = 0

Parabola, Finding the Vertex :

 6.2      Find the Vertex of   y = x²+5x+8

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 1 , is positive (greater than zero). 

 Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions. 

 Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex. 

 For any parabola,Ax²+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is  -2.5000  

 Plugging into the parabola formula  -2.5000  for  x  we can calculate the  y -coordinate : 
  y = 1.0 * -2.50 * -2.50 + 5.0 * -2.50 + 8.0
or   y = 1.750

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x²+5x+8
Axis of Symmetry (dashed)  {x}={-2.50} 
Vertex at  {x,y} = {-2.50, 1.75} 
Function has no real roots

Solve Quadratic Equation by Completing The Square

 6.3     Solving   x²+5x+8 = 0 by Completing The Square .

 Subtract  8  from both side of the equation :
   x²+5x = -8

Now the clever bit: Take the coefficient of  x , which is  5 , divide by two, giving  5/2 , and finally square it giving  25/4 

Add  25/4  to both sides of the equation :
  On the right hand side we have :
   -8  +  25/4    or,  (-8/1)+(25/4) 
  The common denominator of the two fractions is  4   Adding  (-32/4)+(25/4)  gives  -7/4 
  So adding to both sides we finally get :
   x²+5x+(25/4) = -7/4

Adding  25/4  has completed the left hand side into a perfect square :
   x²+5x+(25/4)  =
   (x+(5/2)) • (x+(5/2))  =
  (x+(5/2))²
Things which are equal to the same thing are also equal to one another. Since
   x²+5x+(25/4) = -7/4 and
   x²+5x+(25/4) = (x+(5/2))²
then, according to the law of transitivity,
   (x+(5/2))² = -7/4

We'll refer to this Equation as  Eq. #6.3.1  

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
   (x+(5/2))²   is
   (x+(5/2))^2/2 =
  (x+(5/2))¹ =
   x+(5/2)

Now, applying the Square Root Principle to  Eq. #6.3.1  we get:
   x+(5/2) = √ -7/4

Subtract  5/2  from both sides to obtain:
   x = -5/2 + √ -7/4
In Math,  i  is called the imaginary unit. It satisfies   i²  =-1. Both   i   and   -i   are the square roots of   -1 


Since a square root has two values, one positive and the other negative
   x² + 5x + 8 = 0
   has two solutions:
  x = -5/2 + √ 7/4 •  i 
   or
  x = -5/2 - √ 7/4 •  i 

Note that  √ 7/4 can be written as
  √ 7  / √ 4   which is √ 7  / 2

Solve Quadratic Equation using the Quadratic Formula

 6.4     Solving    x²+5x+8 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  x  , the solution for   Ax²+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  x =   ————————
                      2A

  In our case,  A   =     1
                      B   =    5
                      C   =   8

Accordingly,  B²  -  4AC   =
                     25 - 32 =
                     -7

Applying the quadratic formula :

               -5 ± √ -7
   x  =    —————
                    2

In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written  (a+b*i) 

Both   i   and   -i   are the square roots of minus 1

Accordingly,√ -7  = 
                    √ 7 • (-1)  =
                    √ 7  • √ -1   =
                    ±  √ 7  • i

  √ 7   , rounded to 4 decimal digits, is   2.6458
 So now we are looking at:
           x  =  ( -5 ±  2.646 i ) / 2

Two imaginary solutions :

 x =(-5+√-7)/2=(-5+i√ 7 )/2= -2.5000+1.3229i
  or: 
 x =(-5-√-7)/2=(-5-i√ 7 )/2= -2.5000-1.3229i

Two solutions were found :

1.  x =(-5-√-7)/2=(-5-i√ 7 )/2= -2.5000-1.3229i
2.  x =(-5+√-7)/2=(-5+i√ 7 )/2= -2.5000+1.3229i