Solution - Adding, subtracting and finding the least common multiple

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "3.6" was replaced by "(36/10)". 3 more similar replacement(s)

Step by step solution :

Step  1  :

            18
 Simplify   ——
            5 

Equation at the end of step  1  :

     15         32     18
  ((———•(x2))-(———•x))-——  = 0 
    100        100     5 

Step  2  :

             8
 Simplify   ——
            25

Equation at the end of step  2  :

     15         8     18
  ((———•(x2))-(——•x))-——  = 0 
    100        25     5 
 Step  3  :
             3
 Simplify   ——
            20

Equation at the end of step  3  :

     3          8x     18
  ((—— • x2) -  ——) -  ——  = 0 
    20          25     5 

Step  4  :

Equation at the end of step  4  :

   3x2    8x     18
  (——— -  ——) -  ——  = 0 
   20     25     5 

Step  5  :

Calculating the Least Common Multiple :

 5.1    Find the Least Common Multiple

      The left denominator is :       20 

      The right denominator is :       25 

      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 = 5

   Right_M = L.C.M / R_Deno = 4

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.      3x2 • 5
   ——————————————————  =   ———————
         L.C.M               100  

   R. Mult. • R. Num.      8x • 4
   ——————————————————  =   ——————
         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:

 3x2 • 5 - (8x • 4)     15x2 - 32x
 ——————————————————  =  ——————————
        100                100    

Equation at the end of step  5  :

  (15x2 - 32x)    18
  ———————————— -  ——  = 0 
      100         5 

Step  6  :

Step  7  :

Pulling out like terms :

 7.1     Pull out like factors :

   15x² - 32x  =   x • (15x - 32) 

Calculating the Least Common Multiple :

 7.2    Find the Least Common Multiple

      The left denominator is :       100 

      The right denominator is :       5 

      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 = 20

Making Equivalent Fractions :

 7.4      Rewrite the two fractions into equivalent fractions

   L. Mult. • L. Num.      x • (15x-32)
   ——————————————————  =   ————————————
         L.C.M                 100     

   R. Mult. • R. Num.      18 • 20
   ——————————————————  =   ———————
         L.C.M               100  

Adding fractions that have a common denominator :

 7.5       Adding up the two equivalent fractions

 x • (15x-32) - (18 • 20)     15x2 - 32x - 360
 ————————————————————————  =  ————————————————
           100                      100       

Trying to factor by splitting the middle term

 7.6     Factoring  15x² - 32x - 360 

The first term is,  15x²  its coefficient is  15 .
The middle term is,  -32x  its coefficient is  -32 .
The last term, "the constant", is  -360 

Step-1 : Multiply the coefficient of the first term by the constant   15 • -360 = -5400 

Step-2 : Find two factors of  -5400  whose sum equals the coefficient of the middle term, which is   -32 .

For tidiness, printing of 42 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  :

  15x2 - 32x - 360
  ————————————————  = 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:

  15x2-32x-360
  ———————————— • 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 :
   15x²-32x-360  = 0

Parabola, Finding the Vertex :

 8.2      Find the Vertex of   y = 15x²-32x-360

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, 15 , 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   1.0667  

 Plugging into the parabola formula   1.0667  for  x  we can calculate the  y -coordinate : 
  y = 15.0 * 1.07 * 1.07 - 32.0 * 1.07 - 360.0
or   y = -377.067

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 15x²-32x-360
Axis of Symmetry (dashed)  {x}={ 1.07} 
Vertex at  {x,y} = { 1.07,-377.07} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-3.95, 0.00} 
Root 2 at  {x,y} = { 6.08, 0.00} 

Solve Quadratic Equation by Completing The Square

 8.3     Solving   15x²-32x-360 = 0 by Completing The Square .

 Divide both sides of the equation by  15  to have 1 as the coefficient of the first term :
   x²-(32/15)x-24 = 0

Add  24  to both side of the equation :
   x²-(32/15)x = 24

Now the clever bit: Take the coefficient of  x , which is  32/15 , divide by two, giving  16/15 , and finally square it giving  256/225 

Add  256/225  to both sides of the equation :
  On the right hand side we have :
   24  +  256/225    or,  (24/1)+(256/225) 
  The common denominator of the two fractions is  225   Adding  (5400/225)+(256/225)  gives  5656/225 
  So adding to both sides we finally get :
   x²-(32/15)x+(256/225) = 5656/225

Adding  256/225  has completed the left hand side into a perfect square :
   x²-(32/15)x+(256/225)  =
   (x-(16/15)) • (x-(16/15))  =
  (x-(16/15))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(32/15)x+(256/225) = 5656/225 and
   x²-(32/15)x+(256/225) = (x-(16/15))²
then, according to the law of transitivity,
   (x-(16/15))² = 5656/225

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
   (x-(16/15))²   is
   (x-(16/15))^2/2 =
  (x-(16/15))¹ =
   x-(16/15)

Now, applying the Square Root Principle to  Eq. #8.3.1  we get:
   x-(16/15) = √ 5656/225

Add  16/15  to both sides to obtain:
   x = 16/15 + √ 5656/225

Since a square root has two values, one positive and the other negative
   x² - (32/15)x - 24 = 0
   has two solutions:
  x = 16/15 + √ 5656/225
   or
  x = 16/15 - √ 5656/225

Note that  √ 5656/225 can be written as
  √ 5656  / √ 225   which is √ 5656  / 15

Solve Quadratic Equation using the Quadratic Formula

 8.4     Solving    15x²-32x-360 = 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   =     15
                      B   =   -32
                      C   =  -360

Accordingly,  B²  -  4AC   =
                     1024 - (-21600) =
                     22624

Applying the quadratic formula :

               32 ± √ 22624
   x  =    ———————
                        30

Can  √ 22624 be simplified ?

Yes!   The prime factorization of  22624   is
   2•2•2•2•2•7•101 
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).

√ 22624   =  √ 2•2•2•2•2•7•101   =2•2•√ 1414   =
                ±  4 • √ 1414

  √ 1414   , rounded to 4 decimal digits, is  37.6032
 So now we are looking at:
           x  =  ( 32 ± 4 •  37.603 ) / 30

Two real solutions:

 x =(32+√22624)/30=(16+2√ 1414 )/15= 6.080

or:

 x =(32-√22624)/30=(16-2√ 1414 )/15= -3.947

Two solutions were found :

1.  x =(32-√22624)/30=(16-2√ 1414 )/15= -3.947
2.  x =(32+√22624)/30=(16+2√ 1414 )/15= 6.080