Solution - Adding, subtracting and finding the least common multiple

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "1.21" was replaced by "(121/100)". 2 more similar replacement(s)

Step by step solution :

Step  1  :

            121
 Simplify   ———
            100

Equation at the end of step  1  :

            22          121
  ((x2) -  (—— • x)) +  ———  > 0 
            10          100

Step  2  :

            11
 Simplify   ——
            5 

Equation at the end of step  2  :

            11          121
  ((x2) -  (—— • x)) +  ———  > 0 
            5           100
 Step  3  :

Rewriting the whole as an Equivalent Fraction :

 3.1   Subtracting a fraction from a whole

Rewrite the whole as a fraction using  5  as the denominator :

           x2     x2 • 5
     x2 =  ——  =  ——————
           1        5   

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 • 5 - (11x)     5x2 - 11x
 ——————————————  =  —————————
       5                5    

Equation at the end of step  3  :

  (5x2 - 11x)    121
  ——————————— +  ———  > 0 
       5         100

Step  4  :

Step  5  :

Pulling out like terms :

 5.1     Pull out like factors :

   5x² - 11x  =   x • (5x - 11) 

Calculating the Least Common Multiple :

 5.2    Find the Least Common Multiple

      The left denominator is :       5 

      The right denominator is :       100 

      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 = 20

   Right_M = L.C.M / R_Deno = 1

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 • (5x-11) • 20
   ——————————————————  =   ————————————————
         L.C.M                   100       

   R. Mult. • R. Num.      121
   ——————————————————  =   ———
         L.C.M             100

Adding fractions that have a common denominator :

 5.5       Adding up the two equivalent fractions

 x • (5x-11) • 20 + 121     100x2 - 220x + 121
 ——————————————————————  =  ——————————————————
          100                      100        

Trying to factor by splitting the middle term

 5.6     Factoring  100x² - 220x + 121 

The first term is,  100x²  its coefficient is  100 .
The middle term is,  -220x  its coefficient is  -220 .
The last term, "the constant", is  +121 

Step-1 : Multiply the coefficient of the first term by the constant   100 • 121 = 12100 

Step-2 : Find two factors of  12100  whose sum equals the coefficient of the middle term, which is   -220 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -110  and  -110 
                     100x² - 110x - 110x - 121

Step-4 : Add up the first 2 terms, pulling out like factors :
                    10x • (10x-11)
              Add up the last 2 terms, pulling out common factors :
                    11 • (10x-11)
Step-5 : Add up the four terms of step 4 :
                    (10x-11)  •  (10x-11)
             Which is the desired factorization

Multiplying Exponential Expressions :

 5.7    Multiply  (10x-11)  by  (10x-11) 

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is  (10x-11)  and the exponents are :
          1 , as  (10x-11)  is the same number as  (10x-11)¹ 
 and   1 , as  (10x-11)  is the same number as  (10x-11)¹ 
The product is therefore,  (10x-11)⁽¹⁺¹⁾ = (10x-11)² 

Equation at the end of step  5  :

  (10x - 11)2
  ———————————  > 0 
      100    

Step  6  :

 6.1    Multiply both sides by  100 

 6.2    Divide both sides by  10  

      (x-(11/10))²   > 0

Solve Basic Inequality :

 6.3      Add  11/10  to both sides

             (x)2  > 11/10

Inequality Plot :

 6.4      Inequality plot for

         0.100 x  - 0.110  >  0

  
 

Supplement : Solving Quadratic Equation Directly

Solving    100x2-220x+121  = 0   directly 

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex :

 7.1      Find the Vertex of   y = 100x²-220x+121

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   1.1000  

 Plugging into the parabola formula   1.1000  for  x  we can calculate the  y -coordinate : 
  y = 100.0 * 1.10 * 1.10 - 220.0 * 1.10 + 121.0
or   y = 0.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 100x²-220x+121
Vertex at  {x,y} = { 1.10, 0.00} 
x-Intercept (Root) :
One Root at  {x,y}={ 1.10, 0.00} 
Note that the root coincides with
the Vertex and the Axis of Symmetry
coinsides with the line  x = 0 

Solve Quadratic Equation by Completing The Square

 7.2     Solving   100x²-220x+121 = 0 by Completing The Square .

 Divide both sides of the equation by  100  to have 1 as the coefficient of the first term :
   x²-(11/5)x+(121/100) = 0

Subtract  121/100  from both side of the equation :
   x²-(11/5)x = -121/100

Now the clever bit: Take the coefficient of  x , which is  11/5 , divide by two, giving  11/10 , and finally square it giving  121/100 

Add  121/100  to both sides of the equation :
  On the right hand side we have :
   -121/100  +  121/100   The common denominator of the two fractions is  100   Adding  (-121/100)+(121/100)  gives  0/100 
  So adding to both sides we finally get :
   x²-(11/5)x+(121/100) = 0

Adding  121/100  has completed the left hand side into a perfect square :
   x²-(11/5)x+(121/100)  =
   (x-(11/10)) • (x-(11/10))  =
  (x-(11/10))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(11/5)x+(121/100) = 0 and
   x²-(11/5)x+(121/100) = (x-(11/10))²
then, according to the law of transitivity,
   (x-(11/10))² = 0

We'll refer to this Equation as  Eq. #7.2.1  

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

Note that the square root of
   (x-(11/10))²   is
   (x-(11/10))^2/2 =
  (x-(11/10))¹ =
   x-(11/10)

Now, applying the Square Root Principle to  Eq. #7.2.1  we get:
   x-(11/10) = √ 0

Add  11/10  to both sides to obtain:
   x = 11/10 + √ 0
The square root of zero is zero

This quadratic equation has one solution only. That's because adding zero is the same as subtracting zero.

The solution is:
   x  =  11/10 

Solve Quadratic Equation using the Quadratic Formula

 7.3     Solving    100x²-220x+121 = 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   =   -220
                      C   =  121

Accordingly,  B²  -  4AC   =
                     48400 - 48400 =
                     0

Applying the quadratic formula :

               220 ± √ 0
   x  =    —————
                    200

The square root of zero is zero

This quadratic equation has one solution only. That's because adding zero is the same as subtracting zero.

The solution is:
  x = 220 / 200 = 1.1000(rounded to 4 decimal places)

One solution was found :