Solution - Adding, subtracting and finding the least common multiple

Step by step solution :

Step  1  :

            131
 Simplify   ———
            30 

Equation at the end of step  1  :

             9          131
  ((x2) -  (—— • x)) +  ———  = 0 
            20          30 

Step  2  :

             9
 Simplify   ——
            20

Equation at the end of step  2  :

             9          131
  ((x2) -  (—— • x)) +  ———  = 0 
            20          30 
 Step  3  :

Rewriting the whole as an Equivalent Fraction :

 3.1   Subtracting a fraction from a whole

Rewrite the whole as a fraction using  20  as the denominator :

           x2     x2 • 20
     x2 =  ——  =  ———————
           1        20   

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 • 20 - (9x)     20x2 - 9x
 ——————————————  =  —————————
       20              20    

Equation at the end of step  3  :

  (20x2 - 9x)    131
  ——————————— +  ———  = 0 
      20         30 

Step  4  :

Step  5  :

Pulling out like terms :

 5.1     Pull out like factors :

   20x² - 9x  =   x • (20x - 9) 

Calculating the Least Common Multiple :

 5.2    Find the Least Common Multiple

      The left denominator is :       20 

      The right denominator is :       30 

      Least Common Multiple:
      60 

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 = 3

   Right_M = L.C.M / R_Deno = 2

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 • (20x-9) • 3
   ——————————————————  =   ———————————————
         L.C.M                   60       

   R. Mult. • R. Num.      131 • 2
   ——————————————————  =   ———————
         L.C.M               60   

Adding fractions that have a common denominator :

 5.5       Adding up the two equivalent fractions

 x • (20x-9) • 3 + 131 • 2     60x2 - 27x + 262
 —————————————————————————  =  ————————————————
            60                        60       

Trying to factor by splitting the middle term

 5.6     Factoring  60x² - 27x + 262 

The first term is,  60x²  its coefficient is  60 .
The middle term is,  -27x  its coefficient is  -27 .
The last term, "the constant", is  +262 

Step-1 : Multiply the coefficient of the first term by the constant   60 • 262 = 15720 

Step-2 : Find two factors of  15720  whose sum equals the coefficient of the middle term, which is   -27 .

For tidiness, printing of 58 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  :

  60x2 - 27x + 262
  ————————————————  = 0 
         60       

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:

  60x2-27x+262
  ———————————— • 60 = 0 • 60
       60     

Now, on the left hand side, the  60  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   60x²-27x+262  = 0

Parabola, Finding the Vertex :

 6.2      Find the Vertex of   y = 60x²-27x+262

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, 60 , 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   0.2250  

 Plugging into the parabola formula   0.2250  for  x  we can calculate the  y -coordinate : 
  y = 60.0 * 0.23 * 0.23 - 27.0 * 0.23 + 262.0
or   y = 258.962

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 60x²-27x+262
Axis of Symmetry (dashed)  {x}={ 0.23} 
Vertex at  {x,y} = { 0.23,258.96} 
Function has no real roots

Solve Quadratic Equation by Completing The Square

 6.3     Solving   60x²-27x+262 = 0 by Completing The Square .

 Divide both sides of the equation by  60  to have 1 as the coefficient of the first term :
   x²-(9/20)x+(131/30) = 0

Subtract  131/30  from both side of the equation :
   x²-(9/20)x = -131/30

Now the clever bit: Take the coefficient of  x , which is  9/20 , divide by two, giving  9/40 , and finally square it giving  81/1600 

Add  81/1600  to both sides of the equation :
  On the right hand side we have :
   -131/30  +  81/1600   The common denominator of the two fractions is  4800   Adding  (-20960/4800)+(243/4800)  gives  -20717/4800 
  So adding to both sides we finally get :
   x²-(9/20)x+(81/1600) = -20717/4800

Adding  81/1600  has completed the left hand side into a perfect square :
   x²-(9/20)x+(81/1600)  =
   (x-(9/40)) • (x-(9/40))  =
  (x-(9/40))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(9/20)x+(81/1600) = -20717/4800 and
   x²-(9/20)x+(81/1600) = (x-(9/40))²
then, according to the law of transitivity,
   (x-(9/40))² = -20717/4800

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-(9/40))²   is
   (x-(9/40))^2/2 =
  (x-(9/40))¹ =
   x-(9/40)

Now, applying the Square Root Principle to  Eq. #6.3.1  we get:
   x-(9/40) = √ -20717/4800

Add  9/40  to both sides to obtain:
   x = 9/40 + √ -20717/4800
In Math,  i  is called the imaginary unit. It satisfies   i²  =-1. Both   i   and   -i   are the square roots of   -1 


Since a square root has two values, one positive and the other negative
   x² - (9/20)x + (131/30) = 0
   has two solutions:
  x = 9/40 + √ 20717/4800 •  i 
   or
  x = 9/40 - √ 20717/4800 •  i 

Note that  √ 20717/4800 can be written as
  √ 20717  / √ 4800 

Solve Quadratic Equation using the Quadratic Formula

 6.4     Solving    60x²-27x+262 = 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   =     60
                      B   =   -27
                      C   =  262

Accordingly,  B²  -  4AC   =
                     729 - 62880 =
                     -62151

Applying the quadratic formula :

               27 ± √ -62151
   x  =    ————————
                        120

In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written  (a+b*i) 

Both   i   and   -i   are the square roots of minus 1

Accordingly,√ -62151  = 
                    √ 62151 • (-1)  =
                    √ 62151  • √ -1   =
                    ±  √ 62151  • i

  √ 62151   , rounded to 4 decimal digits, is  249.3010
 So now we are looking at:
           x  =  ( 27 ±  249.301 i ) / 120

Two imaginary solutions :

 x =(27+√-62151)/120=9/40+i/120√ 62151 = 0.2250+2.0775i
  or: 
 x =(27-√-62151)/120=9/40-i/120√ 62151 = 0.2250-2.0775i

Two solutions were found :

1.  x =(27-√-62151)/120=9/40-i/120√ 62151 = 0.2250-2.0775i
2.  x =(27+√-62151)/120=9/40+i/120√ 62151 = 0.2250+2.0775i