Solution - Adding, subtracting and finding the least common multiple

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "2.7" was replaced by "(27/10)". 3 more similar replacement(s)

Step by step solution :

Step  1  :

            27
 Simplify   ——
            10

Equation at the end of step  1  :

    491        382     27
  ((———•(t2))+(———•t))-——  = 0 
    100        100     10

Step  2  :

            191
 Simplify   ———
            50 

Equation at the end of step  2  :

    491        191     27
  ((———•(t2))+(———•t))-——  = 0 
    100        50      10
 Step  3  :
            491
 Simplify   ———
            100

Equation at the end of step  3  :

    491          191t     27
  ((——— • t2) +  ————) -  ——  = 0 
    100           50      10

Step  4  :

Equation at the end of step  4  :

   491t2    191t     27
  (————— +  ————) -  ——  = 0 
    100      50      10

Step  5  :

Calculating the Least Common Multiple :

 5.1    Find the Least Common Multiple

      The left denominator is :       100 

      The right denominator is :       50 

      Least Common Multiple:
      100 

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 = 1

   Right_M = L.C.M / R_Deno = 2

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.      491t2
   ——————————————————  =   —————
         L.C.M              100 

   R. Mult. • R. Num.      191t • 2
   ——————————————————  =   ————————
         L.C.M               100   

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:

 491t2 + 191t • 2     491t2 + 382t
 ————————————————  =  ————————————
       100                100     

Equation at the end of step  5  :

  (491t2 + 382t)    27
  —————————————— -  ——  = 0 
       100          10

Step  6  :

Step  7  :

Pulling out like terms :

 7.1     Pull out like factors :

   491t² + 382t  =   t • (491t + 382) 

Calculating the Least Common Multiple :

 7.2    Find the Least Common Multiple

      The left denominator is :       100 

      The right denominator is :       10 

      Least Common Multiple:
      100 

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 = 10

Making Equivalent Fractions :

 7.4      Rewrite the two fractions into equivalent fractions

   L. Mult. • L. Num.      t • (491t+382)
   ——————————————————  =   ——————————————
         L.C.M                  100      

   R. Mult. • R. Num.      27 • 10
   ——————————————————  =   ———————
         L.C.M               100  

Adding fractions that have a common denominator :

 7.5       Adding up the two equivalent fractions

 t • (491t+382) - (27 • 10)     491t2 + 382t - 270
 ——————————————————————————  =  ——————————————————
            100                        100        

Trying to factor by splitting the middle term

 7.6     Factoring  491t² + 382t - 270 

The first term is,  491t²  its coefficient is  491 .
The middle term is,  +382t  its coefficient is  382 .
The last term, "the constant", is  -270 

Step-1 : Multiply the coefficient of the first term by the constant   491 • -270 = -132570 

Step-2 : Find two factors of  -132570  whose sum equals the coefficient of the middle term, which is   382 .

For tidiness, printing of 26 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  :

  491t2 + 382t - 270
  ——————————————————  = 0 
         100        

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:

  491t2+382t-270
  —————————————— • 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 :
   491t²+382t-270  = 0

Parabola, Finding the Vertex :

 8.2      Find the Vertex of   y = 491t²+382t-270

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, 491 , 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  -0.3890  

 Plugging into the parabola formula  -0.3890  for  t  we can calculate the  y -coordinate : 
  y = 491.0 * -0.39 * -0.39 + 382.0 * -0.39 - 270.0
or   y = -344.299

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 491t²+382t-270
Axis of Symmetry (dashed)  {t}={-0.39} 
Vertex at  {t,y} = {-0.39,-344.30} 
 t -Intercepts (Roots) :
Root 1 at  {t,y} = {-1.23, 0.00} 
Root 2 at  {t,y} = { 0.45, 0.00} 

Solve Quadratic Equation by Completing The Square

 8.3     Solving   491t²+382t-270 = 0 by Completing The Square .

 Divide both sides of the equation by  491  to have 1 as the coefficient of the first term :
   t²+(382/491)t-(270/491) = 0

Add  270/491  to both side of the equation :
   t²+(382/491)t = 270/491

Now the clever bit: Take the coefficient of  t , which is  382/491 , divide by two, giving  191/491 , and finally square it giving  36481/241081 

Add  36481/241081  to both sides of the equation :
  On the right hand side we have :
   270/491  +  36481/241081   The common denominator of the two fractions is  241081   Adding  (132570/241081)+(36481/241081)  gives  169051/241081 
  So adding to both sides we finally get :
   t²+(382/491)t+(36481/241081) = 169051/241081

Adding  36481/241081  has completed the left hand side into a perfect square :
   t²+(382/491)t+(36481/241081)  =
   (t+(191/491)) • (t+(191/491))  =
  (t+(191/491))²
Things which are equal to the same thing are also equal to one another. Since
   t²+(382/491)t+(36481/241081) = 169051/241081 and
   t²+(382/491)t+(36481/241081) = (t+(191/491))²
then, according to the law of transitivity,
   (t+(191/491))² = 169051/241081

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+(191/491))²   is
   (t+(191/491))^2/2 =
  (t+(191/491))¹ =
   t+(191/491)

Now, applying the Square Root Principle to  Eq. #8.3.1  we get:
   t+(191/491) = √ 169051/241081

Subtract  191/491  from both sides to obtain:
   t = -191/491 + √ 169051/241081

Since a square root has two values, one positive and the other negative
   t² + (382/491)t - (270/491) = 0
   has two solutions:
  t = -191/491 + √ 169051/241081
   or
  t = -191/491 - √ 169051/241081

Note that  √ 169051/241081 can be written as
  √ 169051  / √ 241081   which is √ 169051  / 491

Solve Quadratic Equation using the Quadratic Formula

 8.4     Solving    491t²+382t-270 = 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   =    491
                      B   =   382
                      C   =  -270

Accordingly,  B²  -  4AC   =
                     145924 - (-530280) =
                     676204

Applying the quadratic formula :

               -382 ± √ 676204
   t  =    —————————
                          982

Can  √ 676204 be simplified ?

Yes!   The prime factorization of  676204   is
   2•2•71•2381 
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).

√ 676204   =  √ 2•2•71•2381   =
                ±  2 • √ 169051

  √ 169051   , rounded to 4 decimal digits, is  411.1581
 So now we are looking at:
           t  =  ( -382 ± 2 •  411.158 ) / 982

Two real solutions:

 t =(-382+√676204)/982=(-191+√ 169051 )/491= 0.448

or:

 t =(-382-√676204)/982=(-191-√ 169051 )/491= -1.226

Two solutions were found :

1.  t =(-382-√676204)/982=(-191-√ 169051 )/491= -1.226
2.  t =(-382+√676204)/982=(-191+√ 169051 )/491= 0.448