Solution - Adding, subtracting and finding the least common multiple

Step by step solution :

Step  1  :

            3
 Simplify   —
            4

Equation at the end of step  1  :

    7        2     3
  ((—•(z2))+(—•z))-—  = 0 
    5        7     4

Step  2  :

            2
 Simplify   —
            7

Equation at the end of step  2  :

    7        2     3
  ((—•(z2))+(—•z))-—  = 0 
    5        7     4
 Step  3  :
            7
 Simplify   —
            5

Equation at the end of step  3  :

    7          2z     3
  ((— • z2) +  ——) -  —  = 0 
    5          7      4

Step  4  :

Equation at the end of step  4  :

   7z2    2z     3
  (——— +  ——) -  —  = 0 
    5     7      4

Step  5  :

Calculating the Least Common Multiple :

 5.1    Find the Least Common Multiple

      The left denominator is :       5 

      The right denominator is :       7 

      Least Common Multiple:
      35 

Calculating Multipliers :

 5.2    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 = 7

   Right_M = L.C.M / R_Deno = 5

Making Equivalent Fractions :

 5.3      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.      7z2 • 7
   ——————————————————  =   ———————
         L.C.M               35   

   R. Mult. • R. Num.      2z • 5
   ——————————————————  =   ——————
         L.C.M               35  

Adding fractions that have a common denominator :

 5.4       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:

 7z2 • 7 + 2z • 5     49z2 + 10z
 ————————————————  =  ——————————
        35                35    

Equation at the end of step  5  :

  (49z2 + 10z)    3
  ———————————— -  —  = 0 
       35         4

Step  6  :

Step  7  :

Pulling out like terms :

 7.1     Pull out like factors :

   49z² + 10z  =   z • (49z + 10) 

Calculating the Least Common Multiple :

 7.2    Find the Least Common Multiple

      The left denominator is :       35 

      The right denominator is :       4 

      Least Common Multiple:
      140 

Calculating Multipliers :

 7.3    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 = 4

   Right_M = L.C.M / R_Deno = 35

Making Equivalent Fractions :

 7.4      Rewrite the two fractions into equivalent fractions

   L. Mult. • L. Num.      z • (49z+10) • 4
   ——————————————————  =   ————————————————
         L.C.M                   140       

   R. Mult. • R. Num.      3 • 35
   ——————————————————  =   ——————
         L.C.M              140  

Adding fractions that have a common denominator :

 7.5       Adding up the two equivalent fractions

 z • (49z+10) • 4 - (3 • 35)     196z2 + 40z - 105
 ———————————————————————————  =  —————————————————
             140                        140       

Trying to factor by splitting the middle term

 7.6     Factoring  196z² + 40z - 105 

The first term is,  196z²  its coefficient is  196 .
The middle term is,  +40z  its coefficient is  40 .
The last term, "the constant", is  -105 

Step-1 : Multiply the coefficient of the first term by the constant   196 • -105 = -20580 

Step-2 : Find two factors of  -20580  whose sum equals the coefficient of the middle term, which is   40 .

For tidiness, printing of 42 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  7  :

  196z2 + 40z - 105
  —————————————————  = 0 
         140       

Step  8  :

When a fraction equals zero :

 8.1    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:

  196z2+40z-105
  ————————————— • 140 = 0 • 140
       140     

Now, on the left hand side, the  140  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   196z²+40z-105  = 0

Parabola, Finding the Vertex :

 8.2      Find the Vertex of   y = 196z²+40z-105

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, 196 , 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,Az²+Bz+C,the  z -coordinate of the vertex is given by  -B/(2A) . In our case the  z  coordinate is  -0.1020  

 Plugging into the parabola formula  -0.1020  for  z  we can calculate the  y -coordinate : 
  y = 196.0 * -0.10 * -0.10 + 40.0 * -0.10 - 105.0
or   y = -107.041

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 196z²+40z-105
Axis of Symmetry (dashed)  {z}={-0.10} 
Vertex at  {z,y} = {-0.10,-107.04} 
 z -Intercepts (Roots) :
Root 1 at  {z,y} = {-0.84, 0.00} 
Root 2 at  {z,y} = { 0.64, 0.00} 

Solve Quadratic Equation by Completing The Square

 8.3     Solving   196z²+40z-105 = 0 by Completing The Square .

 Divide both sides of the equation by  196  to have 1 as the coefficient of the first term :
   z²+(10/49)z-(15/28) = 0

Add  15/28  to both side of the equation :
   z²+(10/49)z = 15/28

Now the clever bit: Take the coefficient of  z , which is  10/49 , divide by two, giving  5/49 , and finally square it giving  25/2401 

Add  25/2401  to both sides of the equation :
  On the right hand side we have :
   15/28  +  25/2401   The common denominator of the two fractions is  9604   Adding  (5145/9604)+(100/9604)  gives  5245/9604 
  So adding to both sides we finally get :
   z²+(10/49)z+(25/2401) = 5245/9604

Adding  25/2401  has completed the left hand side into a perfect square :
   z²+(10/49)z+(25/2401)  =
   (z+(5/49)) • (z+(5/49))  =
  (z+(5/49))²
Things which are equal to the same thing are also equal to one another. Since
   z²+(10/49)z+(25/2401) = 5245/9604 and
   z²+(10/49)z+(25/2401) = (z+(5/49))²
then, according to the law of transitivity,
   (z+(5/49))² = 5245/9604

We'll refer to this Equation as  Eq. #8.3.1  

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

Note that the square root of
   (z+(5/49))²   is
   (z+(5/49))^2/2 =
  (z+(5/49))¹ =
   z+(5/49)

Now, applying the Square Root Principle to  Eq. #8.3.1  we get:
   z+(5/49) = √ 5245/9604

Subtract  5/49  from both sides to obtain:
   z = -5/49 + √ 5245/9604

Since a square root has two values, one positive and the other negative
   z² + (10/49)z - (15/28) = 0
   has two solutions:
  z = -5/49 + √ 5245/9604
   or
  z = -5/49 - √ 5245/9604

Note that  √ 5245/9604 can be written as
  √ 5245  / √ 9604   which is √ 5245  / 98

Solve Quadratic Equation using the Quadratic Formula

 8.4     Solving    196z²+40z-105 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  z  , the solution for   Az²+Bz+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  z =   ————————
                      2A

  In our case,  A   =    196
                      B   =    40
                      C   =  -105

Accordingly,  B²  -  4AC   =
                     1600 - (-82320) =
                     83920

Applying the quadratic formula :

               -40 ± √ 83920
   z  =    ————————
                        392

Can  √ 83920 be simplified ?

Yes!   The prime factorization of  83920   is
   2•2•2•2•5•1049 
To be able to remove something from under the radical, there have to be  2  instances of it (because we are taking a square i.e. second root).

√ 83920   =  √ 2•2•2•2•5•1049   =2•2•√ 5245   =
                ±  4 • √ 5245

  √ 5245   , rounded to 4 decimal digits, is  72.4224
 So now we are looking at:
           z  =  ( -40 ± 4 •  72.422 ) / 392

Two real solutions:

 z =(-40+√83920)/392=-5/49+1/98√ 5245 = 0.637

or:

 z =(-40-√83920)/392=-5/49-1/98√ 5245 = -0.841

Two solutions were found :

1.  z =(-40-√83920)/392=-5/49-1/98√ 5245 = -0.841
2.  z =(-40+√83920)/392=-5/49+1/98√ 5245 = 0.637