Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((0 -  (2•7x2)) +  1035x) -  10266  = 0 

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   -14x² + 1035x - 10266  =   -1 • (14x² - 1035x + 10266) 

Trying to factor by splitting the middle term

 3.2     Factoring  14x² - 1035x + 10266 

The first term is,  14x²  its coefficient is  14 .
The middle term is,  -1035x  its coefficient is  -1035 .
The last term, "the constant", is  +10266 

Step-1 : Multiply the coefficient of the first term by the constant   14 • 10266 = 143724 

Step-2 : Find two factors of  143724  whose sum equals the coefficient of the middle term, which is   -1035 .

For tidiness, printing of 90 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  :

  -14x2 + 1035x - 10266  = 0 

Step  4  :

Parabola, Finding the Vertex :

 4.1      Find the Vertex of   y = -14x²+1035x-10266

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, -14 , 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  36.9643  

 Plugging into the parabola formula  36.9643  for  x  we can calculate the  y -coordinate : 
  y = -14.0 * 36.96 * 36.96 + 1035.0 * 36.96 - 10266.0
or   y = 8863.018

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = -14x²+1035x-10266
Axis of Symmetry (dashed)  {x}={36.96} 
Vertex at  {x,y} = {36.96,8863.02} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {62.13, 0.00} 
Root 2 at  {x,y} = {11.80, 0.00} 

Solve Quadratic Equation by Completing The Square

 4.2     Solving   -14x²+1035x-10266 = 0 by Completing The Square .

 Multiply both sides of the equation by  (-1)  to obtain positive coefficient for the first term:
 14x²-1035x+10266 = 0  Divide both sides of the equation by  14  to have 1 as the coefficient of the first term :
   x²-(1035/14)x+(5133/7) = 0

Subtract  5133/7  from both side of the equation :
   x²-(1035/14)x = -5133/7

Now the clever bit: Take the coefficient of  x , which is  1035/14 , divide by two, giving  1035/28 , and finally square it giving  1035/28 

Add  1035/28  to both sides of the equation :
  On the right hand side we have :
   -5133/7  +  1035/28   The common denominator of the two fractions is  28   Adding  (-20532/28)+(1035/28)  gives  -19497/28 
  So adding to both sides we finally get :
   x²-(1035/14)x+(1035/28) = -19497/28

Adding  1035/28  has completed the left hand side into a perfect square :
   x²-(1035/14)x+(1035/28)  =
   (x-(1035/28)) • (x-(1035/28))  =
  (x-(1035/28))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(1035/14)x+(1035/28) = -19497/28 and
   x²-(1035/14)x+(1035/28) = (x-(1035/28))²
then, according to the law of transitivity,
   (x-(1035/28))² = -19497/28

We'll refer to this Equation as  Eq. #4.2.1  

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

Note that the square root of
   (x-(1035/28))²   is
   (x-(1035/28))^2/2 =
  (x-(1035/28))¹ =
   x-(1035/28)

Now, applying the Square Root Principle to  Eq. #4.2.1  we get:
   x-(1035/28) = √ -19497/28

Add  1035/28  to both sides to obtain:
   x = 1035/28 + √ -19497/28
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 - (1035/14)x + (5133/7) = 0
   has two solutions:
  x = 1035/28 + √ 19497/28 •  i 
   or
  x = 1035/28 - √ 19497/28 •  i 

Note that  √ 19497/28 can be written as
  √ 19497  / √ 28 

Solve Quadratic Equation using the Quadratic Formula

 4.3     Solving    -14x²+1035x-10266 = 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   =    -14
                      B   =   1035
                      C   =  -10266

Accordingly,  B²  -  4AC   =
                     1071225 - 574896 =
                     496329

Applying the quadratic formula :

               -1035 ± √ 496329
   x  =    ——————————
                           -28

  √ 496329   , rounded to 4 decimal digits, is  704.5062
 So now we are looking at:
           x  =  ( -1035 ±  704.506 ) / -28

Two real solutions:

 x =(-1035+√496329)/-28=11.803

or:

 x =(-1035-√496329)/-28=62.125

Two solutions were found :

1.  x =(-1035-√496329)/-28=62.125
2.  x =(-1035+√496329)/-28=11.803