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 :

                     5/x-2-20/x^2-4-(1)=0 

Step by step solution :

Step  1  :

            20
 Simplify   ——
            x2

Equation at the end of step  1  :

     5          20           
  (((— -  2) -  ——) -  4) -  1  = 0 
     x          x2           

Step  2  :

            5
 Simplify   —
            x

Equation at the end of step  2  :

     5          20           
  (((— -  2) -  ——) -  4) -  1  = 0 
     x          x2           

Step  3  :

Rewriting the whole as an Equivalent Fraction :

 3.1   Subtracting a whole from a fraction

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

         2     2 • x
    2 =  —  =  —————
         1       x  

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 :

 3.2       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:

 5 - (2 • x)     5 - 2x
 ———————————  =  ——————
      x            x   

Equation at the end of step  3  :

    (5 - 2x)    20           
  ((———————— -  ——) -  4) -  1  = 0 
       x        x2           

Step  4  :

Calculating the Least Common Multiple :

 4.1    Find the Least Common Multiple

      The left denominator is :       x 

      The right denominator is :       x² 

      Least Common Multiple:
      x² 

Calculating Multipliers :

 4.2    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

   Right_M = L.C.M / R_Deno = 1

Making Equivalent Fractions :

 4.3      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.      (5-2x) • x
   ——————————————————  =   ——————————
         L.C.M                 x2    

   R. Mult. • R. Num.      20
   ——————————————————  =   ——
         L.C.M             x2

Adding fractions that have a common denominator :

 4.4       Adding up the two equivalent fractions

 (5-2x) • x - (20)     -2x2 + 5x - 20
 —————————————————  =  ——————————————
        x2                   x2      

Equation at the end of step  4  :

   (-2x2 + 5x - 20)          
  (———————————————— -  4) -  1  = 0 
          x2                 

Step  5  :

Rewriting the whole as an Equivalent Fraction :

 5.1   Subtracting a whole from a fraction

Rewrite the whole as a fraction using  x²  as the denominator :

         4     4 • x2
    4 =  —  =  ——————
         1       x2  

Step  6  :

Pulling out like terms :

 6.1     Pull out like factors :

   -2x² + 5x - 20  =   -1 • (2x² - 5x + 20) 

Trying to factor by splitting the middle term

 6.2     Factoring  2x² - 5x + 20 

The first term is,  2x²  its coefficient is  2 .
The middle term is,  -5x  its coefficient is  -5 .
The last term, "the constant", is  +20 

Step-1 : Multiply the coefficient of the first term by the constant   2 • 20 = 40 

Step-2 : Find two factors of  40  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

Adding fractions that have a common denominator :

 6.3       Adding up the two equivalent fractions

 (-2x2+5x-20) - (4 • x2)     -6x2 + 5x - 20
 ———————————————————————  =  ——————————————
           x2                      x2      

Equation at the end of step  6  :

  (-6x2 + 5x - 20)    
  ———————————————— -  1  = 0 
         x2           

Step  7  :

Rewriting the whole as an Equivalent Fraction :

 7.1   Subtracting a whole from a fraction

Rewrite the whole as a fraction using  x²  as the denominator :

         1     1 • x2
    1 =  —  =  ——————
         1       x2  

Step  8  :

Pulling out like terms :

 8.1     Pull out like factors :

   -6x² + 5x - 20  =   -1 • (6x² - 5x + 20) 

Trying to factor by splitting the middle term

 8.2     Factoring  6x² - 5x + 20 

The first term is,  6x²  its coefficient is  6 .
The middle term is,  -5x  its coefficient is  -5 .
The last term, "the constant", is  +20 

Step-1 : Multiply the coefficient of the first term by the constant   6 • 20 = 120 

Step-2 : Find two factors of  120  whose sum equals the coefficient of the middle term, which is   -5 .

For tidiness, printing of 26 lines which failed to find two such factors, was suppressed

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

Adding fractions that have a common denominator :

 8.3       Adding up the two equivalent fractions

 (-6x2+5x-20) - (x2)     -7x2 + 5x - 20
 ———————————————————  =  ——————————————
         x2                    x2      

Step  9  :

Pulling out like terms :

 9.1     Pull out like factors :

   -7x² + 5x - 20  =   -1 • (7x² - 5x + 20) 

Trying to factor by splitting the middle term

 9.2     Factoring  7x² - 5x + 20 

The first term is,  7x²  its coefficient is  7 .
The middle term is,  -5x  its coefficient is  -5 .
The last term, "the constant", is  +20 

Step-1 : Multiply the coefficient of the first term by the constant   7 • 20 = 140 

Step-2 : Find two factors of  140  whose sum equals the coefficient of the middle term, which is   -5 .

For tidiness, printing of 18 lines which failed to find two such factors, was suppressed

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

Equation at the end of step  9  :

  -7x2 + 5x - 20
  ——————————————  = 0 
        x2      

Step  10  :

When a fraction equals zero :

 10.1    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:

  -7x2+5x-20
  —————————— • x2 = 0 • x2
      x2    

Now, on the left hand side, the  x²  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   -7x²+5x-20  = 0

Parabola, Finding the Vertex :

 10.2      Find the Vertex of   y = -7x²+5x-20

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens down and accordingly has a highest point (AKA absolute maximum) .    We know this even before plotting  "y"  because the coefficient of the first term, -7 , is negative (smaller 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   0.3571  

 Plugging into the parabola formula   0.3571  for  x  we can calculate the  y -coordinate : 
  y = -7.0 * 0.36 * 0.36 + 5.0 * 0.36 - 20.0
or   y = -19.107

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = -7x²+5x-20
Axis of Symmetry (dashed)  {x}={ 0.36} 
Vertex at  {x,y} = { 0.36,-19.11} 
Function has no real roots

Solve Quadratic Equation by Completing The Square

 10.3     Solving   -7x²+5x-20 = 0 by Completing The Square .

 Multiply both sides of the equation by  (-1)  to obtain positive coefficient for the first term:
 7x²-5x+20 = 0  Divide both sides of the equation by  7  to have 1 as the coefficient of the first term :
   x²-(5/7)x+(20/7) = 0

Subtract  20/7  from both side of the equation :
   x²-(5/7)x = -20/7

Now the clever bit: Take the coefficient of  x , which is  5/7 , divide by two, giving  5/14 , and finally square it giving  25/196 

Add  25/196  to both sides of the equation :
  On the right hand side we have :
   -20/7  +  25/196   The common denominator of the two fractions is  196   Adding  (-560/196)+(25/196)  gives  -535/196 
  So adding to both sides we finally get :
   x²-(5/7)x+(25/196) = -535/196

Adding  25/196  has completed the left hand side into a perfect square :
   x²-(5/7)x+(25/196)  =
   (x-(5/14)) • (x-(5/14))  =
  (x-(5/14))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(5/7)x+(25/196) = -535/196 and
   x²-(5/7)x+(25/196) = (x-(5/14))²
then, according to the law of transitivity,
   (x-(5/14))² = -535/196

We'll refer to this Equation as  Eq. #10.3.1  

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

Note that the square root of
   (x-(5/14))²   is
   (x-(5/14))^2/2 =
  (x-(5/14))¹ =
   x-(5/14)

Now, applying the Square Root Principle to  Eq. #10.3.1  we get:
   x-(5/14) = √ -535/196

Add  5/14  to both sides to obtain:
   x = 5/14 + √ -535/196
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 - (5/7)x + (20/7) = 0
   has two solutions:
  x = 5/14 + √ 535/196 •  i 
   or
  x = 5/14 - √ 535/196 •  i 

Note that  √ 535/196 can be written as
  √ 535  / √ 196   which is √ 535  / 14

Solve Quadratic Equation using the Quadratic Formula

 10.4     Solving    -7x²+5x-20 = 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   =     -7
                      B   =    5
                      C   =  -20

Accordingly,  B²  -  4AC   =
                     25 - 560 =
                     -535

Applying the quadratic formula :

               -5 ± √ -535
   x  =    ——————
                      -14

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,√ -535  = 
                    √ 535 • (-1)  =
                    √ 535  • √ -1   =
                    ±  √ 535  • i

  √ 535   , rounded to 4 decimal digits, is  23.1301
 So now we are looking at:
           x  =  ( -5 ±  23.130 i ) / -14

Two imaginary solutions :

 x =(-5+√-535)/-14=(5-i√ 535 )/14= 0.3571+1.6521i
  or: 
 x =(-5-√-535)/-14=(5+i√ 535 )/14= 0.3571-1.6521i

Two solutions were found :

1.  x =(-5-√-535)/-14=(5+i√ 535 )/14= 0.3571-1.6521i
2.  x =(-5+√-535)/-14=(5-i√ 535 )/14= 0.3571+1.6521i