Solution - Adding, subtracting and finding the least common multiple

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "6.52" was replaced by "(652/100)". 2 more similar replacement(s)

Step by step solution :

Step  1  :

            163
 Simplify   ———
            25 

Equation at the end of step  1  :

        11         163
  ((0-(———•(x2)))+(———•x))+2  = 0 
       100         25 
 Step  2  :
             11
 Simplify   ———
            100

Equation at the end of step  2  :

           11           163x     
  ((0 -  (——— • x2)) +  ————) +  2  = 0 
          100            25      

Step  3  :

Equation at the end of step  3  :

         11x2     163x     
  ((0 -  ————) +  ————) +  2  = 0 
         100       25      

Step  4  :

Calculating the Least Common Multiple :

 4.1    Find the Least Common Multiple

      The left denominator is :       100 

      The right denominator is :       25 

      Least Common Multiple:
      100 

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 = 1

   Right_M = L.C.M / R_Deno = 4

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.      -11x2
   ——————————————————  =   —————
         L.C.M              100 

   R. Mult. • R. Num.      163x • 4
   ——————————————————  =   ————————
         L.C.M               100   

Adding fractions that have a common denominator :

 4.4       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:

 -11x2 + 163x • 4     652x - 11x2
 ————————————————  =  ———————————
       100                100    

Equation at the end of step  4  :

  (652x - 11x2)    
  ————————————— +  2  = 0 
       100         

Step  5  :

Rewriting the whole as an Equivalent Fraction :

 5.1   Adding a whole to a fraction

Rewrite the whole as a fraction using  100  as the denominator :

         2     2 • 100
    2 =  —  =  ———————
         1       100  

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

Step  6  :

Pulling out like terms :

 6.1     Pull out like factors :

   652x - 11x²  =   -x • (11x - 652) 

Adding fractions that have a common denominator :

 6.2       Adding up the two equivalent fractions

 -x • (11x-652) + 2 • 100     -11x2 + 652x + 200
 ————————————————————————  =  ——————————————————
           100                       100        

Trying to factor by splitting the middle term

 6.3     Factoring  -11x² + 652x + 200 

The first term is,  -11x²  its coefficient is  -11 .
The middle term is,  +652x  its coefficient is  652 .
The last term, "the constant", is  +200 

Step-1 : Multiply the coefficient of the first term by the constant   -11 • 200 = -2200 

Step-2 : Find two factors of  -2200  whose sum equals the coefficient of the middle term, which is   652 .

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  6  :

  -11x2 + 652x + 200
  ——————————————————  = 0 
         100        

Step  7  :

When a fraction equals zero :

 7.1    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:

  -11x2+652x+200
  —————————————— • 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 :
   -11x²+652x+200  = 0

Parabola, Finding the Vertex :

 7.2      Find the Vertex of   y = -11x²+652x+200

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, -11 , 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  29.6364  

 Plugging into the parabola formula  29.6364  for  x  we can calculate the  y -coordinate : 
  y = -11.0 * 29.64 * 29.64 + 652.0 * 29.64 + 200.0
or   y = 9861.455

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = -11x²+652x+200
Axis of Symmetry (dashed)  {x}={29.64} 
Vertex at  {x,y} = {29.64,9861.45} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {59.58, 0.00} 
Root 2 at  {x,y} = {-0.31, 0.00} 

Solve Quadratic Equation by Completing The Square

 7.3     Solving   -11x²+652x+200 = 0 by Completing The Square .

 Multiply both sides of the equation by  (-1)  to obtain positive coefficient for the first term:
 11x²-652x-200 = 0  Divide both sides of the equation by  11  to have 1 as the coefficient of the first term :
   x²-(652/11)x-(200/11) = 0

Add  200/11  to both side of the equation :
   x²-(652/11)x = 200/11

Now the clever bit: Take the coefficient of  x , which is  652/11 , divide by two, giving  326/11 , and finally square it giving  106276/121 

Add  106276/121  to both sides of the equation :
  On the right hand side we have :
   200/11  +  106276/121   The common denominator of the two fractions is  121   Adding  (2200/121)+(106276/121)  gives  108476/121 
  So adding to both sides we finally get :
   x²-(652/11)x+(106276/121) = 108476/121

Adding  106276/121  has completed the left hand side into a perfect square :
   x²-(652/11)x+(106276/121)  =
   (x-(326/11)) • (x-(326/11))  =
  (x-(326/11))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(652/11)x+(106276/121) = 108476/121 and
   x²-(652/11)x+(106276/121) = (x-(326/11))²
then, according to the law of transitivity,
   (x-(326/11))² = 108476/121

We'll refer to this Equation as  Eq. #7.3.1  

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

Note that the square root of
   (x-(326/11))²   is
   (x-(326/11))^2/2 =
  (x-(326/11))¹ =
   x-(326/11)

Now, applying the Square Root Principle to  Eq. #7.3.1  we get:
   x-(326/11) = √ 108476/121

Add  326/11  to both sides to obtain:
   x = 326/11 + √ 108476/121

Since a square root has two values, one positive and the other negative
   x² - (652/11)x - (200/11) = 0
   has two solutions:
  x = 326/11 + √ 108476/121
   or
  x = 326/11 - √ 108476/121

Note that  √ 108476/121 can be written as
  √ 108476  / √ 121   which is √ 108476  / 11

Solve Quadratic Equation using the Quadratic Formula

 7.4     Solving    -11x²+652x+200 = 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   =    -11
                      B   =   652
                      C   =  200

Accordingly,  B²  -  4AC   =
                     425104 - (-8800) =
                     433904

Applying the quadratic formula :

               -652 ± √ 433904
   x  =    —————————
                          -22

Can  √ 433904 be simplified ?

Yes!   The prime factorization of  433904   is
   2•2•2•2•47•577 
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).

√ 433904   =  √ 2•2•2•2•47•577   =2•2•√ 27119   =
                ±  4 • √ 27119

  √ 27119   , rounded to 4 decimal digits, is  164.6785
 So now we are looking at:
           x  =  ( -652 ± 4 •  164.678 ) / -22

Two real solutions:

 x =(-652+√433904)/-22=(326-2√ 27119 )/11= -0.305

or:

 x =(-652-√433904)/-22=(326+2√ 27119 )/11= 59.578

Two solutions were found :

1.  x =(-652-√433904)/-22=(326+2√ 27119 )/11= 59.578
2.  x =(-652+√433904)/-22=(326-2√ 27119 )/11= -0.305