Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((2•23x2) -  200x) +  200  = 0 

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   46x² - 200x + 200  =   2 • (23x² - 100x + 100) 

Trying to factor by splitting the middle term

 3.2     Factoring  23x² - 100x + 100 

The first term is,  23x²  its coefficient is  23 .
The middle term is,  -100x  its coefficient is  -100 .
The last term, "the constant", is  +100 

Step-1 : Multiply the coefficient of the first term by the constant   23 • 100 = 2300 

Step-2 : Find two factors of  2300  whose sum equals the coefficient of the middle term, which is   -100 .

For tidiness, printing of 30 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  3  :

  2 • (23x2 - 100x + 100)  = 0 

Step  4  :

Equations which are never true :

 4.1      Solve :    2   =  0

This equation has no solution.
A a non-zero constant never equals zero.

Parabola, Finding the Vertex :

 4.2      Find the Vertex of   y = 23x²-100x+100

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, 23 , 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   2.1739  

 Plugging into the parabola formula   2.1739  for  x  we can calculate the  y -coordinate : 
  y = 23.0 * 2.17 * 2.17 - 100.0 * 2.17 + 100.0
or   y = -8.696

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 23x²-100x+100
Axis of Symmetry (dashed)  {x}={ 2.17} 
Vertex at  {x,y} = { 2.17,-8.70} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = { 1.56, 0.00} 
Root 2 at  {x,y} = { 2.79, 0.00} 

Solve Quadratic Equation by Completing The Square

 4.3     Solving   23x²-100x+100 = 0 by Completing The Square .

 Divide both sides of the equation by  23  to have 1 as the coefficient of the first term :
   x²-(100/23)x+(100/23) = 0

Subtract  100/23  from both side of the equation :
   x²-(100/23)x = -100/23

Now the clever bit: Take the coefficient of  x , which is  100/23 , divide by two, giving  50/23 , and finally square it giving  2500/529 

Add  2500/529  to both sides of the equation :
  On the right hand side we have :
   -100/23  +  2500/529   The common denominator of the two fractions is  529   Adding  (-2300/529)+(2500/529)  gives  200/529 
  So adding to both sides we finally get :
   x²-(100/23)x+(2500/529) = 200/529

Adding  2500/529  has completed the left hand side into a perfect square :
   x²-(100/23)x+(2500/529)  =
   (x-(50/23)) • (x-(50/23))  =
  (x-(50/23))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(100/23)x+(2500/529) = 200/529 and
   x²-(100/23)x+(2500/529) = (x-(50/23))²
then, according to the law of transitivity,
   (x-(50/23))² = 200/529

We'll refer to this Equation as  Eq. #4.3.1  

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
   (x-(50/23))²   is
   (x-(50/23))^2/2 =
  (x-(50/23))¹ =
   x-(50/23)

Now, applying the Square Root Principle to  Eq. #4.3.1  we get:
   x-(50/23) = √ 200/529

Add  50/23  to both sides to obtain:
   x = 50/23 + √ 200/529

Since a square root has two values, one positive and the other negative
   x² - (100/23)x + (100/23) = 0
   has two solutions:
  x = 50/23 + √ 200/529
   or
  x = 50/23 - √ 200/529

Note that  √ 200/529 can be written as
  √ 200  / √ 529   which is √ 200  / 23

Solve Quadratic Equation using the Quadratic Formula

 4.4     Solving    23x²-100x+100 = 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   =     23
                      B   =   -100
                      C   =  100

Accordingly,  B²  -  4AC   =
                     10000 - 9200 =
                     800

Applying the quadratic formula :

               100 ± √ 800
   x  =    ——————
                      46

Can  √ 800 be simplified ?

Yes!   The prime factorization of  800   is
   2•2•2•2•2•5•5 
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).

√ 800   =  √ 2•2•2•2•2•5•5   =2•2•5•√ 2   =
                ±  20 • √ 2

  √ 2   , rounded to 4 decimal digits, is   1.4142
 So now we are looking at:
           x  =  ( 100 ± 20 •  1.414 ) / 46

Two real solutions:

 x =(100+√800)/46=(50+10√ 2 )/23= 2.789

or:

 x =(100-√800)/46=(50-10√ 2 )/23= 1.559

Two solutions were found :

1.  x =(100-√800)/46=(50-10√ 2 )/23= 1.559
2.  x =(100+√800)/46=(50+10√ 2 )/23= 2.789