Solution - Adding, subtracting and finding the least common multiple

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "5.25" was replaced by "(525/100)". 3 more similar replacement(s)

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

           (583/100)*x^2-(619/100)*x-((525/100))=0 

Step by step solution :

Step  1  :

            21
 Simplify   ——
            4 

Equation at the end of step  1  :

    583        619     21
  ((———•(x2))-(———•x))-——  = 0 
    100        100     4 

Step  2  :

            619
 Simplify   ———
            100

Equation at the end of step  2  :

    583        619     21
  ((———•(x2))-(———•x))-——  = 0 
    100        100     4 
 Step  3  :
            583
 Simplify   ———
            100

Equation at the end of step  3  :

    583          619x     21
  ((——— • x2) -  ————) -  ——  = 0 
    100          100      4 

Step  4  :

Equation at the end of step  4  :

   583x2    619x     21
  (————— -  ————) -  ——  = 0 
    100     100      4 

Step  5  :

Adding fractions which have a common denominator :

 5.1       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:

 583x2 - (619x)     583x2 - 619x
 ——————————————  =  ————————————
      100               100     

Equation at the end of step  5  :

  (583x2 - 619x)    21
  —————————————— -  ——  = 0 
       100          4 

Step  6  :

Step  7  :

Pulling out like terms :

 7.1     Pull out like factors :

   583x² - 619x  =   x • (583x - 619) 

Calculating the Least Common Multiple :

 7.2    Find the Least Common Multiple

      The left denominator is :       100 

      The right denominator is :       4 

      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 = 25

Making Equivalent Fractions :

 7.4      Rewrite the two fractions into equivalent fractions

   L. Mult. • L. Num.      x • (583x-619)
   ——————————————————  =   ——————————————
         L.C.M                  100      

   R. Mult. • R. Num.      21 • 25
   ——————————————————  =   ———————
         L.C.M               100  

Adding fractions that have a common denominator :

 7.5       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:

 x • (583x-619) - (21 • 25)     583x2 - 619x - 525
 ——————————————————————————  =  ——————————————————
            100                        100        

Trying to factor by splitting the middle term

 7.6     Factoring  583x² - 619x - 525 

The first term is,  583x²  its coefficient is  583 .
The middle term is,  -619x  its coefficient is  -619 .
The last term, "the constant", is  -525 

Step-1 : Multiply the coefficient of the first term by the constant   583 • -525 = -306075 

Step-2 : Find two factors of  -306075  whose sum equals the coefficient of the middle term, which is   -619 .

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  :

  583x2 - 619x - 525
  ——————————————————  = 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:

  583x2-619x-525
  —————————————— • 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 :
   583x²-619x-525  = 0

Parabola, Finding the Vertex :

 8.2      Find the Vertex of   y = 583x²-619x-525

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, 583 , 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   0.5309  

 Plugging into the parabola formula   0.5309  for  x  we can calculate the  y -coordinate : 
  y = 583.0 * 0.53 * 0.53 - 619.0 * 0.53 - 525.0
or   y = -689.306

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 583x²-619x-525
Axis of Symmetry (dashed)  {x}={ 0.53} 
Vertex at  {x,y} = { 0.53,-689.31} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-0.56, 0.00} 
Root 2 at  {x,y} = { 1.62, 0.00} 

Solve Quadratic Equation by Completing The Square

 8.3     Solving   583x²-619x-525 = 0 by Completing The Square .

 Divide both sides of the equation by  583  to have 1 as the coefficient of the first term :
   x²-(619/583)x-(525/583) = 0

Add  525/583  to both side of the equation :
   x²-(619/583)x = 525/583

Now the clever bit: Take the coefficient of  x , which is  619/583 , divide by two, giving  619/1166 , and finally square it giving  619/1166 

Add  619/1166  to both sides of the equation :
  On the right hand side we have :
   525/583  +  619/1166   The common denominator of the two fractions is  1166   Adding  (1050/1166)+(619/1166)  gives  1669/1166 
  So adding to both sides we finally get :
   x²-(619/583)x+(619/1166) = 1669/1166

Adding  619/1166  has completed the left hand side into a perfect square :
   x²-(619/583)x+(619/1166)  =
   (x-(619/1166)) • (x-(619/1166))  =
  (x-(619/1166))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(619/583)x+(619/1166) = 1669/1166 and
   x²-(619/583)x+(619/1166) = (x-(619/1166))²
then, according to the law of transitivity,
   (x-(619/1166))² = 1669/1166

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-(619/1166))²   is
   (x-(619/1166))^2/2 =
  (x-(619/1166))¹ =
   x-(619/1166)

Now, applying the Square Root Principle to  Eq. #8.3.1  we get:
   x-(619/1166) = √ 1669/1166

Add  619/1166  to both sides to obtain:
   x = 619/1166 + √ 1669/1166

Since a square root has two values, one positive and the other negative
   x² - (619/583)x - (525/583) = 0
   has two solutions:
  x = 619/1166 + √ 1669/1166
   or
  x = 619/1166 - √ 1669/1166

Note that  √ 1669/1166 can be written as
  √ 1669  / √ 1166 

Solve Quadratic Equation using the Quadratic Formula

 8.4     Solving    583x²-619x-525 = 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   =    583
                      B   =   -619
                      C   =  -525

Accordingly,  B²  -  4AC   =
                     383161 - (-1224300) =
                     1607461

Applying the quadratic formula :

               619 ± √ 1607461
   x  =    —————————
                          1166

  √ 1607461   , rounded to 4 decimal digits, is  1267.8569
 So now we are looking at:
           x  =  ( 619 ±  1267.857 ) / 1166

Two real solutions:

 x =(619+√1607461)/1166= 1.618

or:

 x =(619-√1607461)/1166=-0.556

Two solutions were found :

1.  x =(619-√1607461)/1166=-0.556
2.  x =(619+√1607461)/1166= 1.618