Solution - Adding, subtracting and finding the least common multiple

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "0.027" was replaced by "(027/1000)". 3 more similar replacement(s)

Step by step solution :

Step  1  :

             27 
 Simplify   ————
            1000

Equation at the end of step  1  :

    24         54      27 
  ((——•(x2))-(———•x))-————  = 0 
    10        100     1000

Step  2  :

            27
 Simplify   ——
            50

Equation at the end of step  2  :

    24        27      27 
  ((——•(x2))-(——•x))-————  = 0 
    10        50     1000
 Step  3  :
            12
 Simplify   ——
            5 

Equation at the end of step  3  :

    12          27x      27 
  ((—— • x2) -  ———) -  ————  = 0 
    5           50      1000

Step  4  :

Equation at the end of step  4  :

   12x2    27x      27 
  (———— -  ———) -  ————  = 0 
    5      50      1000

Step  5  :

Calculating the Least Common Multiple :

 5.1    Find the Least Common Multiple

      The left denominator is :       5 

      The right denominator is :       50 

      Least Common Multiple:
      50 

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 = 10

   Right_M = L.C.M / R_Deno = 1

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.      12x2 • 10
   ——————————————————  =   —————————
         L.C.M                50    

   R. Mult. • R. Num.      27x
   ——————————————————  =   ———
         L.C.M             50 

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:

 12x2 • 10 - (27x)     120x2 - 27x
 —————————————————  =  ———————————
        50                 50     

Equation at the end of step  5  :

  (120x2 - 27x)     27 
  ————————————— -  ————  = 0 
       50          1000

Step  6  :

Step  7  :

Pulling out like terms :

 7.1     Pull out like factors :

   120x² - 27x  =   3x • (40x - 9) 

Calculating the Least Common Multiple :

 7.2    Find the Least Common Multiple

      The left denominator is :       50 

      The right denominator is :       1000 

      Least Common Multiple:
      1000 

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 = 20

   Right_M = L.C.M / R_Deno = 1

Making Equivalent Fractions :

 7.4      Rewrite the two fractions into equivalent fractions

   L. Mult. • L. Num.      3x • (40x-9) • 20
   ——————————————————  =   —————————————————
         L.C.M                   1000       

   R. Mult. • R. Num.       27 
   ——————————————————  =   ————
         L.C.M             1000

Adding fractions that have a common denominator :

 7.5       Adding up the two equivalent fractions

 3x • (40x-9) • 20 - (27)     2400x2 - 540x - 27
 ————————————————————————  =  ——————————————————
           1000                      1000       

Step  8  :

Pulling out like terms :

 8.1     Pull out like factors :

   2400x² - 540x - 27  =   3 • (800x² - 180x - 9) 

Trying to factor by splitting the middle term

 8.2     Factoring  800x² - 180x - 9 

The first term is,  800x²  its coefficient is  800 .
The middle term is,  -180x  its coefficient is  -180 .
The last term, "the constant", is  -9 

Step-1 : Multiply the coefficient of the first term by the constant   800 • -9 = -7200 

Step-2 : Find two factors of  -7200  whose sum equals the coefficient of the middle term, which is   -180 .

For tidiness, printing of 48 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  8  :

  3 • (800x2 - 180x - 9)
  ——————————————————————  = 0 
           1000         

Step  9  :

When a fraction equals zero :

 9.1    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:

  3•(800x2-180x-9)
  ———————————————— • 1000 = 0 • 1000
        1000      

Now, on the left hand side, the  1000  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   3  •  (800x²-180x-9)  = 0

Equations which are never true :

 9.2      Solve :    3   =  0

This equation has no solution.
A a non-zero constant never equals zero.

Parabola, Finding the Vertex :

 9.3      Find the Vertex of   y = 800x²-180x-9

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, 800 , 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   0.1125  

 Plugging into the parabola formula   0.1125  for  x  we can calculate the  y -coordinate : 
  y = 800.0 * 0.11 * 0.11 - 180.0 * 0.11 - 9.0
or   y = -19.125

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 800x²-180x-9
Axis of Symmetry (dashed)  {x}={ 0.11} 
Vertex at  {x,y} = { 0.11,-19.12} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-0.04, 0.00} 
Root 2 at  {x,y} = { 0.27, 0.00} 

Solve Quadratic Equation by Completing The Square

 9.4     Solving   800x²-180x-9 = 0 by Completing The Square .

 Divide both sides of the equation by  800  to have 1 as the coefficient of the first term :
   x²-(9/40)x-(9/800) = 0

Add  9/800  to both side of the equation :
   x²-(9/40)x = 9/800

Now the clever bit: Take the coefficient of  x , which is  9/40 , divide by two, giving  9/80 , and finally square it giving  81/6400 

Add  81/6400  to both sides of the equation :
  On the right hand side we have :
   9/800  +  81/6400   The common denominator of the two fractions is  6400   Adding  (72/6400)+(81/6400)  gives  153/6400 
  So adding to both sides we finally get :
   x²-(9/40)x+(81/6400) = 153/6400

Adding  81/6400  has completed the left hand side into a perfect square :
   x²-(9/40)x+(81/6400)  =
   (x-(9/80)) • (x-(9/80))  =
  (x-(9/80))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(9/40)x+(81/6400) = 153/6400 and
   x²-(9/40)x+(81/6400) = (x-(9/80))²
then, according to the law of transitivity,
   (x-(9/80))² = 153/6400

We'll refer to this Equation as  Eq. #9.4.1  

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

Note that the square root of
   (x-(9/80))²   is
   (x-(9/80))^2/2 =
  (x-(9/80))¹ =
   x-(9/80)

Now, applying the Square Root Principle to  Eq. #9.4.1  we get:
   x-(9/80) = √ 153/6400

Add  9/80  to both sides to obtain:
   x = 9/80 + √ 153/6400

Since a square root has two values, one positive and the other negative
   x² - (9/40)x - (9/800) = 0
   has two solutions:
  x = 9/80 + √ 153/6400
   or
  x = 9/80 - √ 153/6400

Note that  √ 153/6400 can be written as
  √ 153  / √ 6400   which is √ 153  / 80

Solve Quadratic Equation using the Quadratic Formula

 9.5     Solving    800x²-180x-9 = 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   =    800
                      B   =   -180
                      C   =   -9

Accordingly,  B²  -  4AC   =
                     32400 - (-28800) =
                     61200

Applying the quadratic formula :

               180 ± √ 61200
   x  =    ————————
                        1600

Can  √ 61200 be simplified ?

Yes!   The prime factorization of  61200   is
   2•2•2•2•3•3•5•5•17 
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).

√ 61200   =  √ 2•2•2•2•3•3•5•5•17   =2•2•3•5•√ 17   =
                ±  60 • √ 17

  √ 17   , rounded to 4 decimal digits, is   4.1231
 So now we are looking at:
           x  =  ( 180 ± 60 •  4.123 ) / 1600

Two real solutions:

 x =(180+√61200)/1600=(9+3√ 17 )/80= 0.267

or:

 x =(180-√61200)/1600=(9-3√ 17 )/80= -0.042

Two solutions were found :

1.  x =(180-√61200)/1600=(9-3√ 17 )/80= -0.042
2.  x =(180+√61200)/1600=(9+3√ 17 )/80= 0.267