Solution - Adding, subtracting and finding the least common multiple

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "10.9" was replaced by "(109/10)". 3 more similar replacement(s)

Step by step solution :

Step  1  :

            109
 Simplify   ———
            10 

Equation at the end of step  1  :

    49        1705     109
  ((——•(t2))-(————•t))-———  = 0 
    10        100      10 

Step  2  :

            341
 Simplify   ———
            20 

Equation at the end of step  2  :

    49        341     109
  ((——•(t2))-(———•t))-———  = 0 
    10        20      10 
 Step  3  :
            49
 Simplify   ——
            10

Equation at the end of step  3  :

    49          341t     109
  ((—— • t2) -  ————) -  ———  = 0 
    10           20      10 

Step  4  :

Equation at the end of step  4  :

   49t2    341t     109
  (———— -  ————) -  ———  = 0 
    10      20      10 

Step  5  :

Calculating the Least Common Multiple :

 5.1    Find the Least Common Multiple

      The left denominator is :       10 

      The right denominator is :       20 

      Least Common Multiple:
      20 

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 = 2

   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.      49t2 • 2
   ——————————————————  =   ————————
         L.C.M                20   

   R. Mult. • R. Num.      341t
   ——————————————————  =   ————
         L.C.M              20 

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:

 49t2 • 2 - (341t)     98t2 - 341t
 —————————————————  =  ———————————
        20                 20     

Equation at the end of step  5  :

  (98t2 - 341t)    109
  ————————————— -  ———  = 0 
       20          10 

Step  6  :

Step  7  :

Pulling out like terms :

 7.1     Pull out like factors :

   98t² - 341t  =   t • (98t - 341) 

Calculating the Least Common Multiple :

 7.2    Find the Least Common Multiple

      The left denominator is :       20 

      The right denominator is :       10 

      Least Common Multiple:
      20 

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 = 1

   Right_M = L.C.M / R_Deno = 2

Making Equivalent Fractions :

 7.4      Rewrite the two fractions into equivalent fractions

   L. Mult. • L. Num.      t • (98t-341)
   ——————————————————  =   —————————————
         L.C.M                  20      

   R. Mult. • R. Num.      109 • 2
   ——————————————————  =   ———————
         L.C.M               20   

Adding fractions that have a common denominator :

 7.5       Adding up the two equivalent fractions

 t • (98t-341) - (109 • 2)     98t2 - 341t - 218
 —————————————————————————  =  —————————————————
            20                        20        

Trying to factor by splitting the middle term

 7.6     Factoring  98t² - 341t - 218 

The first term is,  98t²  its coefficient is  98 .
The middle term is,  -341t  its coefficient is  -341 .
The last term, "the constant", is  -218 

Step-1 : Multiply the coefficient of the first term by the constant   98 • -218 = -21364 

Step-2 : Find two factors of  -21364  whose sum equals the coefficient of the middle term, which is   -341 .

For tidiness, printing of 12 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  7  :

  98t2 - 341t - 218
  —————————————————  = 0 
         20        

Step  8  :

When a fraction equals zero :

 8.1    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:

  98t2-341t-218
  ————————————— • 20 = 0 • 20
       20      

Now, on the left hand side, the  20  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   98t²-341t-218  = 0

Parabola, Finding the Vertex :

 8.2      Find the Vertex of   y = 98t²-341t-218

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, 98 , 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,At²+Bt+C,the  t -coordinate of the vertex is given by  -B/(2A) . In our case the  t  coordinate is   1.7398  

 Plugging into the parabola formula   1.7398  for  t  we can calculate the  y -coordinate : 
  y = 98.0 * 1.74 * 1.74 - 341.0 * 1.74 - 218.0
or   y = -514.635

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 98t²-341t-218
Axis of Symmetry (dashed)  {t}={ 1.74} 
Vertex at  {t,y} = { 1.74,-514.64} 
 t -Intercepts (Roots) :
Root 1 at  {t,y} = {-0.55, 0.00} 
Root 2 at  {t,y} = { 4.03, 0.00} 

Solve Quadratic Equation by Completing The Square

 8.3     Solving   98t²-341t-218 = 0 by Completing The Square .

 Divide both sides of the equation by  98  to have 1 as the coefficient of the first term :
   t²-(341/98)t-(109/49) = 0

Add  109/49  to both side of the equation :
   t²-(341/98)t = 109/49

Now the clever bit: Take the coefficient of  t , which is  341/98 , divide by two, giving  341/196 , and finally square it giving  116281/38416 

Add  116281/38416  to both sides of the equation :
  On the right hand side we have :
   109/49  +  116281/38416   The common denominator of the two fractions is  38416   Adding  (85456/38416)+(116281/38416)  gives  201737/38416 
  So adding to both sides we finally get :
   t²-(341/98)t+(116281/38416) = 201737/38416

Adding  116281/38416  has completed the left hand side into a perfect square :
   t²-(341/98)t+(116281/38416)  =
   (t-(341/196)) • (t-(341/196))  =
  (t-(341/196))²
Things which are equal to the same thing are also equal to one another. Since
   t²-(341/98)t+(116281/38416) = 201737/38416 and
   t²-(341/98)t+(116281/38416) = (t-(341/196))²
then, according to the law of transitivity,
   (t-(341/196))² = 201737/38416

We'll refer to this Equation as  Eq. #8.3.1  

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

Note that the square root of
   (t-(341/196))²   is
   (t-(341/196))^2/2 =
  (t-(341/196))¹ =
   t-(341/196)

Now, applying the Square Root Principle to  Eq. #8.3.1  we get:
   t-(341/196) = √ 201737/38416

Add  341/196  to both sides to obtain:
   t = 341/196 + √ 201737/38416

Since a square root has two values, one positive and the other negative
   t² - (341/98)t - (109/49) = 0
   has two solutions:
  t = 341/196 + √ 201737/38416
   or
  t = 341/196 - √ 201737/38416

Note that  √ 201737/38416 can be written as
  √ 201737  / √ 38416   which is √ 201737  / 196

Solve Quadratic Equation using the Quadratic Formula

 8.4     Solving    98t²-341t-218 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  t  , the solution for   At²+Bt+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  t =   ————————
                      2A

  In our case,  A   =     98
                      B   =   -341
                      C   =  -218

Accordingly,  B²  -  4AC   =
                     116281 - (-85456) =
                     201737

Applying the quadratic formula :

               341 ± √ 201737
   t  =    ————————
                        196

  √ 201737   , rounded to 4 decimal digits, is  449.1514
 So now we are looking at:
           t  =  ( 341 ±  449.151 ) / 196

Two real solutions:

 t =(341+√201737)/196= 4.031

or:

 t =(341-√201737)/196=-0.552

Two solutions were found :

1.  t =(341-√201737)/196=-0.552
2.  t =(341+√201737)/196= 4.031