Solution - Adding, subtracting and finding the least common multiple

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "3.36" was replaced by "(336/100)". 2 more similar replacement(s)

Step by step solution :

Step  1  :

            84
 Simplify   ——
            25

Equation at the end of step  1  :

            535          84
  ((x2) +  (——— • x)) -  ——  = 0 
            100          25

Step  2  :

            107
 Simplify   ———
            20 

Equation at the end of step  2  :

            107          84
  ((x2) +  (——— • x)) -  ——  = 0 
            20           25
 Step  3  :

Rewriting the whole as an Equivalent Fraction :

 3.1   Adding a fraction to a whole

Rewrite the whole as a fraction using  20  as the denominator :

           x2     x2 • 20
     x2 =  ——  =  ———————
           1        20   

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:

 x2 • 20 + 107x     20x2 + 107x
 ——————————————  =  ———————————
       20               20     

Equation at the end of step  3  :

  (20x2 + 107x)    84
  ————————————— -  ——  = 0 
       20          25

Step  4  :

Step  5  :

Pulling out like terms :

 5.1     Pull out like factors :

   20x² + 107x  =   x • (20x + 107) 

Calculating the Least Common Multiple :

 5.2    Find the Least Common Multiple

      The left denominator is :       20 

      The right denominator is :       25 

      Least Common Multiple:
      100 

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 = 5

   Right_M = L.C.M / R_Deno = 4

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.      x • (20x+107) • 5
   ——————————————————  =   —————————————————
         L.C.M                    100       

   R. Mult. • R. Num.      84 • 4
   ——————————————————  =   ——————
         L.C.M              100  

Adding fractions that have a common denominator :

 5.5       Adding up the two equivalent fractions

 x • (20x+107) • 5 - (84 • 4)     100x2 + 535x - 336
 ————————————————————————————  =  ——————————————————
             100                         100        

Trying to factor by splitting the middle term

 5.6     Factoring  100x² + 535x - 336 

The first term is,  100x²  its coefficient is  100 .
The middle term is,  +535x  its coefficient is  535 .
The last term, "the constant", is  -336 

Step-1 : Multiply the coefficient of the first term by the constant   100 • -336 = -33600 

Step-2 : Find two factors of  -33600  whose sum equals the coefficient of the middle term, which is   535 .

For tidiness, printing of 78 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  5  :

  100x2 + 535x - 336
  ——————————————————  = 0 
         100        

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:

  100x2+535x-336
  —————————————— • 100 = 0 • 100
       100      

Now, on the left hand side, the  100  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   100x²+535x-336  = 0

Parabola, Finding the Vertex :

 6.2      Find the Vertex of   y = 100x²+535x-336

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, 100 , 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  -2.6750  

 Plugging into the parabola formula  -2.6750  for  x  we can calculate the  y -coordinate : 
  y = 100.0 * -2.67 * -2.67 + 535.0 * -2.67 - 336.0
or   y = -1051.562

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 100x²+535x-336
Axis of Symmetry (dashed)  {x}={-2.67} 
Vertex at  {x,y} = {-2.67,-1051.56} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-5.92, 0.00} 
Root 2 at  {x,y} = { 0.57, 0.00} 

Solve Quadratic Equation by Completing The Square

 6.3     Solving   100x²+535x-336 = 0 by Completing The Square .

 Divide both sides of the equation by  100  to have 1 as the coefficient of the first term :
   x²+(107/20)x-(84/25) = 0

Add  84/25  to both side of the equation :
   x²+(107/20)x = 84/25

Now the clever bit: Take the coefficient of  x , which is  107/20 , divide by two, giving  107/40 , and finally square it giving  11449/1600 

Add  11449/1600  to both sides of the equation :
  On the right hand side we have :
   84/25  +  11449/1600   The common denominator of the two fractions is  1600   Adding  (5376/1600)+(11449/1600)  gives  16825/1600 
  So adding to both sides we finally get :
   x²+(107/20)x+(11449/1600) = 673/64

Adding  11449/1600  has completed the left hand side into a perfect square :
   x²+(107/20)x+(11449/1600)  =
   (x+(107/40)) • (x+(107/40))  =
  (x+(107/40))²
Things which are equal to the same thing are also equal to one another. Since
   x²+(107/20)x+(11449/1600) = 673/64 and
   x²+(107/20)x+(11449/1600) = (x+(107/40))²
then, according to the law of transitivity,
   (x+(107/40))² = 673/64

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+(107/40))²   is
   (x+(107/40))^2/2 =
  (x+(107/40))¹ =
   x+(107/40)

Now, applying the Square Root Principle to  Eq. #6.3.1  we get:
   x+(107/40) = √ 673/64

Subtract  107/40  from both sides to obtain:
   x = -107/40 + √ 673/64

Since a square root has two values, one positive and the other negative
   x² + (107/20)x - (84/25) = 0
   has two solutions:
  x = -107/40 + √ 673/64
   or
  x = -107/40 - √ 673/64

Note that  √ 673/64 can be written as
  √ 673  / √ 64   which is √ 673  / 8

Solve Quadratic Equation using the Quadratic Formula

 6.4     Solving    100x²+535x-336 = 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   =    100
                      B   =   535
                      C   =  -336

Accordingly,  B²  -  4AC   =
                     286225 - (-134400) =
                     420625

Applying the quadratic formula :

               -535 ± √ 420625
   x  =    —————————
                          200

Can  √ 420625 be simplified ?

Yes!   The prime factorization of  420625   is
   5•5•5•5•673 
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).

√ 420625   =  √ 5•5•5•5•673   =5•5•√ 673   =
                ±  25 • √ 673

  √ 673   , rounded to 4 decimal digits, is  25.9422
 So now we are looking at:
           x  =  ( -535 ± 25 •  25.942 ) / 200

Two real solutions:

 x =(-535+√420625)/200=-107/40+1/8√ 673 = 0.568

or:

 x =(-535-√420625)/200=-107/40-1/8√ 673 = -5.918

Two solutions were found :

1.  x =(-535-√420625)/200=-107/40-1/8√ 673 = -5.918
2.  x =(-535+√420625)/200=-107/40+1/8√ 673 = 0.568