Solution - Quadratic equations

Step by step solution :

Step  1  :

             x
 Simplify   ——
            50

Equation at the end of step  1  :

                   x     
  ((0 -  (x2)) +  ——) -  1  = 0 
                  50     

Step  2  :

Rewriting the whole as an Equivalent Fraction :

 2.1   Adding a fraction to a whole

Rewrite the whole as a fraction using  50  as the denominator :

             -x2     -x2 • 50
      -x2 =  ———  =  ————————
              1         50   

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 :

 2.2       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 • 50 + x     x - 50x2
 ————————————  =  ————————
      50             50   

Equation at the end of step  2  :

  (x - 50x2)    
  —————————— -  1  = 0 
      50        

Step  3  :

Rewriting the whole as an Equivalent Fraction :

 3.1   Subtracting a whole from a fraction

Rewrite the whole as a fraction using  50  as the denominator :

         1     1 • 50
    1 =  —  =  ——————
         1       50  

Step  4  :

Pulling out like terms :

 4.1     Pull out like factors :

   x - 50x²  =   -x • (50x - 1) 

Adding fractions that have a common denominator :

 4.2       Adding up the two equivalent fractions

 -x • (50x-1) - (50)     -50x2 + x - 50
 ———————————————————  =  ——————————————
         50                    50      

Step  5  :

Pulling out like terms :

 5.1     Pull out like factors :

   -50x² + x - 50  =   -1 • (50x² - x + 50) 

Trying to factor by splitting the middle term

 5.2     Factoring  50x² - x + 50 

The first term is,  50x²  its coefficient is  50 .
The middle term is,  -x  its coefficient is  -1 .
The last term, "the constant", is  +50 

Step-1 : Multiply the coefficient of the first term by the constant   50 • 50 = 2500 

Step-2 : Find two factors of  2500  whose sum equals the coefficient of the middle term, which is   -1 .

For tidiness, printing of 24 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  :

  -50x2 + x - 50
  ——————————————  = 0 
        50      

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:

  -50x2+x-50
  —————————— • 50 = 0 • 50
      50    

Now, on the left hand side, the  50  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   -50x²+x-50  = 0

Parabola, Finding the Vertex :

 6.2      Find the Vertex of   y = -50x²+x-50

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens down and accordingly has a highest point (AKA absolute maximum) .    We know this even before plotting  "y"  because the coefficient of the first term, -50 , is negative (smaller 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   0.0100  

 Plugging into the parabola formula   0.0100  for  x  we can calculate the  y -coordinate : 
  y = -50.0 * 0.01 * 0.01 + 1.0 * 0.01 - 50.0
or   y = -49.995

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = -50x²+x-50
Axis of Symmetry (dashed)  {x}={ 0.01} 
Vertex at  {x,y} = { 0.01,-49.99} 
Function has no real roots

Solve Quadratic Equation by Completing The Square

 6.3     Solving   -50x²+x-50 = 0 by Completing The Square .

 Multiply both sides of the equation by  (-1)  to obtain positive coefficient for the first term:
 50x²-x+50 = 0  Divide both sides of the equation by  50  to have 1 as the coefficient of the first term :
   x²-(1/50)x+1 = 0

Subtract  1  from both side of the equation :
   x²-(1/50)x = -1

Now the clever bit: Take the coefficient of  x , which is  1/50 , divide by two, giving  1/100 , and finally square it giving  1/10000 

Add  1/10000  to both sides of the equation :
  On the right hand side we have :
   -1  +  1/10000    or,  (-1/1)+(1/10000) 
  The common denominator of the two fractions is  10000   Adding  (-10000/10000)+(1/10000)  gives  -9999/10000 
  So adding to both sides we finally get :
   x²-(1/50)x+(1/10000) = -9999/10000

Adding  1/10000  has completed the left hand side into a perfect square :
   x²-(1/50)x+(1/10000)  =
   (x-(1/100)) • (x-(1/100))  =
  (x-(1/100))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(1/50)x+(1/10000) = -9999/10000 and
   x²-(1/50)x+(1/10000) = (x-(1/100))²
then, according to the law of transitivity,
   (x-(1/100))² = -9999/10000

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-(1/100))²   is
   (x-(1/100))^2/2 =
  (x-(1/100))¹ =
   x-(1/100)

Now, applying the Square Root Principle to  Eq. #6.3.1  we get:
   x-(1/100) = √ -9999/10000

Add  1/100  to both sides to obtain:
   x = 1/100 + √ -9999/10000
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 - (1/50)x + 1 = 0
   has two solutions:
  x = 1/100 + √ 9999/10000 •  i 
   or
  x = 1/100 - √ 9999/10000 •  i 

Note that  √ 9999/10000 can be written as
  √ 9999  / √ 10000   which is √ 9999  / 100

Solve Quadratic Equation using the Quadratic Formula

 6.4     Solving    -50x²+x-50 = 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   =    -50
                      B   =    1
                      C   =  -50

Accordingly,  B²  -  4AC   =
                     1 - 10000 =
                     -9999

Applying the quadratic formula :

               -1 ± √ -9999
   x  =    ———————
                        -100

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,√ -9999  = 
                    √ 9999 • (-1)  =
                    √ 9999  • √ -1   =
                    ±  √ 9999  • i

Can  √ 9999 be simplified ?

Yes!   The prime factorization of  9999   is
   3•3•11•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).

√ 9999   =  √ 3•3•11•101   =
                ±  3 • √ 1111

  √ 1111   , rounded to 4 decimal digits, is  33.3317
 So now we are looking at:
           x  =  ( -1 ± 3 •  33.332 i ) / -100

Two imaginary solutions :

 x =(-1+√-9999)/-100=(1-3i√ 1111 )/100= 0.0100+0.9999i
  or: 
 x =(-1-√-9999)/-100=(1+3i√ 1111 )/100= 0.0100-0.9999i

Two solutions were found :

1.  x =(-1-√-9999)/-100=(1+3i√ 1111 )/100= 0.0100-0.9999i
2.  x =(-1+√-9999)/-100=(1-3i√ 1111 )/100= 0.0100+0.9999i