Solution - Adding, subtracting and finding the least common multiple

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "0.063" was replaced by "(063/1000)". 2 more similar replacement(s)

Step by step solution :

Step  1  :

             63 
 Simplify   ————
            1000

Equation at the end of step  1  :

             11            63 
  ((x2) -  (———— • x)) -  ————  = 0 
            1000          1000

Step  2  :

             11 
 Simplify   ————
            1000

Equation at the end of step  2  :

             11            63 
  ((x2) -  (———— • x)) -  ————  = 0 
            1000          1000
 Step  3  :

Rewriting the whole as an Equivalent Fraction :

 3.1   Subtracting a fraction from a whole

Rewrite the whole as a fraction using  1000  as the denominator :

           x2     x2 • 1000
     x2 =  ——  =  —————————
           1        1000   

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 • 1000 - (11x)     1000x2 - 11x
 —————————————————  =  ————————————
       1000                1000    

Equation at the end of step  3  :

  (1000x2 - 11x)     63 
  —————————————— -  ————  = 0 
       1000         1000

Step  4  :

Step  5  :

Pulling out like terms :

 5.1     Pull out like factors :

   1000x² - 11x  =   x • (1000x - 11) 

Adding fractions which have a common denominator :

 5.2       Adding fractions which have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

 x • (1000x-11) - (63)     1000x2 - 11x - 63
 —————————————————————  =  —————————————————
         1000                    1000       

Trying to factor by splitting the middle term

 5.3     Factoring  1000x² - 11x - 63 

The first term is,  1000x²  its coefficient is  1000 .
The middle term is,  -11x  its coefficient is  -11 .
The last term, "the constant", is  -63 

Step-1 : Multiply the coefficient of the first term by the constant   1000 • -63 = -63000 

Step-2 : Find two factors of  -63000  whose sum equals the coefficient of the middle term, which is   -11 .

For tidiness, printing of 90 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  :

  1000x2 - 11x - 63
  —————————————————  = 0 
        1000       

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:

  1000x2-11x-63
  ————————————— • 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 :
   1000x²-11x-63  = 0

Parabola, Finding the Vertex :

 6.2      Find the Vertex of   y = 1000x²-11x-63

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, 1000 , 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   0.0055  

 Plugging into the parabola formula   0.0055  for  x  we can calculate the  y -coordinate : 
  y = 1000.0 * 0.01 * 0.01 - 11.0 * 0.01 - 63.0
or   y = -63.030

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 1000x²-11x-63
Axis of Symmetry (dashed)  {x}={ 0.01} 
Vertex at  {x,y} = { 0.01,-63.03} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-0.25, 0.00} 
Root 2 at  {x,y} = { 0.26, 0.00} 

Solve Quadratic Equation by Completing The Square

 6.3     Solving   1000x²-11x-63 = 0 by Completing The Square .

 Divide both sides of the equation by  1000  to have 1 as the coefficient of the first term :
   x²-(11/1000)x-(63/1000) = 0

Add  63/1000  to both side of the equation :
   x²-(11/1000)x = 63/1000

Now the clever bit: Take the coefficient of  x , which is  11/1000 , divide by two, giving  11/2000 , and finally square it giving  11/2000 

Add  11/2000  to both sides of the equation :
  On the right hand side we have :
   63/1000  +  11/2000   The common denominator of the two fractions is  2000   Adding  (126/2000)+(11/2000)  gives  137/2000 
  So adding to both sides we finally get :
   x²-(11/1000)x+(11/2000) = 137/2000

Adding  11/2000  has completed the left hand side into a perfect square :
   x²-(11/1000)x+(11/2000)  =
   (x-(11/2000)) • (x-(11/2000))  =
  (x-(11/2000))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(11/1000)x+(11/2000) = 137/2000 and
   x²-(11/1000)x+(11/2000) = (x-(11/2000))²
then, according to the law of transitivity,
   (x-(11/2000))² = 137/2000

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-(11/2000))²   is
   (x-(11/2000))^2/2 =
  (x-(11/2000))¹ =
   x-(11/2000)

Now, applying the Square Root Principle to  Eq. #6.3.1  we get:
   x-(11/2000) = √ 137/2000

Add  11/2000  to both sides to obtain:
   x = 11/2000 + √ 137/2000

Since a square root has two values, one positive and the other negative
   x² - (11/1000)x - (63/1000) = 0
   has two solutions:
  x = 11/2000 + √ 137/2000
   or
  x = 11/2000 - √ 137/2000

Note that  √ 137/2000 can be written as
  √ 137  / √ 2000 

Solve Quadratic Equation using the Quadratic Formula

 6.4     Solving    1000x²-11x-63 = 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   =    1000
                      B   =   -11
                      C   =  -63

Accordingly,  B²  -  4AC   =
                     121 - (-252000) =
                     252121

Applying the quadratic formula :

               11 ± √ 252121
   x  =    ————————
                        2000

  √ 252121   , rounded to 4 decimal digits, is  502.1165
 So now we are looking at:
           x  =  ( 11 ±  502.117 ) / 2000

Two real solutions:

 x =(11+√252121)/2000= 0.257

or:

 x =(11-√252121)/2000=-0.246

Two solutions were found :

1.  x =(11-√252121)/2000=-0.246
2.  x =(11+√252121)/2000= 0.257