Solution - Adding, subtracting and finding the least common multiple

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "5.9" was replaced by "(59/10)". 3 more similar replacement(s)

Step by step solution :

Step  1  :

            59
 Simplify   ——
            10

Equation at the end of step  1  :

    51        64     59
  ((——•(x2))+(——•x))-——  = 0 
    10        10     10

Step  2  :

            32
 Simplify   ——
            5 

Equation at the end of step  2  :

    51        32     59
  ((——•(x2))+(——•x))-——  = 0 
    10        5      10
 Step  3  :
            51
 Simplify   ——
            10

Equation at the end of step  3  :

    51          32x     59
  ((—— • x2) +  ———) -  ——  = 0 
    10           5      10

Step  4  :

Equation at the end of step  4  :

   51x2    32x     59
  (———— +  ———) -  ——  = 0 
    10      5      10

Step  5  :

Calculating the Least Common Multiple :

 5.1    Find the Least Common Multiple

      The left denominator is :       10 

      The right denominator is :       5 

      Least Common Multiple:
      10 

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.      51x2
   ——————————————————  =   ————
         L.C.M              10 

   R. Mult. • R. Num.      32x • 2
   ——————————————————  =   ———————
         L.C.M               10   

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:

 51x2 + 32x • 2     51x2 + 64x
 ——————————————  =  ——————————
       10               10    

Equation at the end of step  5  :

  (51x2 + 64x)    59
  ———————————— -  ——  = 0 
       10         10

Step  6  :

Step  7  :

Pulling out like terms :

 7.1     Pull out like factors :

   51x² + 64x  =   x • (51x + 64) 

Adding fractions which have a common denominator :

 7.2       Adding fractions which have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

 x • (51x+64) - (59)     51x2 + 64x - 59
 ———————————————————  =  ———————————————
         10                    10       

Trying to factor by splitting the middle term

 7.3     Factoring  51x² + 64x - 59 

The first term is,  51x²  its coefficient is  51 .
The middle term is,  +64x  its coefficient is  64 .
The last term, "the constant", is  -59 

Step-1 : Multiply the coefficient of the first term by the constant   51 • -59 = -3009 

Step-2 : Find two factors of  -3009  whose sum equals the coefficient of the middle term, which is   64 .

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Equation at the end of step  7  :

  51x2 + 64x - 59
  ———————————————  = 0 
        10       

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:

  51x2+64x-59
  ——————————— • 10 = 0 • 10
      10     

Now, on the left hand side, the  10  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   51x²+64x-59  = 0

Parabola, Finding the Vertex :

 8.2      Find the Vertex of   y = 51x²+64x-59

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, 51 , 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  -0.6275  

 Plugging into the parabola formula  -0.6275  for  x  we can calculate the  y -coordinate : 
  y = 51.0 * -0.63 * -0.63 + 64.0 * -0.63 - 59.0
or   y = -79.078

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 51x²+64x-59
Axis of Symmetry (dashed)  {x}={-0.63} 
Vertex at  {x,y} = {-0.63,-79.08} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-1.87, 0.00} 
Root 2 at  {x,y} = { 0.62, 0.00} 

Solve Quadratic Equation by Completing The Square

 8.3     Solving   51x²+64x-59 = 0 by Completing The Square .

 Divide both sides of the equation by  51  to have 1 as the coefficient of the first term :
   x²+(64/51)x-(59/51) = 0

Add  59/51  to both side of the equation :
   x²+(64/51)x = 59/51

Now the clever bit: Take the coefficient of  x , which is  64/51 , divide by two, giving  32/51 , and finally square it giving  1024/2601 

Add  1024/2601  to both sides of the equation :
  On the right hand side we have :
   59/51  +  1024/2601   The common denominator of the two fractions is  2601   Adding  (3009/2601)+(1024/2601)  gives  4033/2601 
  So adding to both sides we finally get :
   x²+(64/51)x+(1024/2601) = 4033/2601

Adding  1024/2601  has completed the left hand side into a perfect square :
   x²+(64/51)x+(1024/2601)  =
   (x+(32/51)) • (x+(32/51))  =
  (x+(32/51))²
Things which are equal to the same thing are also equal to one another. Since
   x²+(64/51)x+(1024/2601) = 4033/2601 and
   x²+(64/51)x+(1024/2601) = (x+(32/51))²
then, according to the law of transitivity,
   (x+(32/51))² = 4033/2601

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+(32/51))²   is
   (x+(32/51))^2/2 =
  (x+(32/51))¹ =
   x+(32/51)

Now, applying the Square Root Principle to  Eq. #8.3.1  we get:
   x+(32/51) = √ 4033/2601

Subtract  32/51  from both sides to obtain:
   x = -32/51 + √ 4033/2601

Since a square root has two values, one positive and the other negative
   x² + (64/51)x - (59/51) = 0
   has two solutions:
  x = -32/51 + √ 4033/2601
   or
  x = -32/51 - √ 4033/2601

Note that  √ 4033/2601 can be written as
  √ 4033  / √ 2601   which is √ 4033  / 51

Solve Quadratic Equation using the Quadratic Formula

 8.4     Solving    51x²+64x-59 = 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   =     51
                      B   =    64
                      C   =  -59

Accordingly,  B²  -  4AC   =
                     4096 - (-12036) =
                     16132

Applying the quadratic formula :

               -64 ± √ 16132
   x  =    ————————
                        102

Can  √ 16132 be simplified ?

Yes!   The prime factorization of  16132   is
   2•2•37•109 
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).

√ 16132   =  √ 2•2•37•109   =
                ±  2 • √ 4033

  √ 4033   , rounded to 4 decimal digits, is  63.5059
 So now we are looking at:
           x  =  ( -64 ± 2 •  63.506 ) / 102

Two real solutions:

 x =(-64+√16132)/102=(-32+√ 4033 )/51= 0.618

or:

 x =(-64-√16132)/102=(-32-√ 4033 )/51= -1.873

Two solutions were found :

1.  x =(-64-√16132)/102=(-32-√ 4033 )/51= -1.873
2.  x =(-64+√16132)/102=(-32+√ 4033 )/51= 0.618