Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (11n2 -  6n) -  13  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  11n²-6n-13 

The first term is,  11n²  its coefficient is  11 .
The middle term is,  -6n  its coefficient is  -6 .
The last term, "the constant", is  -13 

Step-1 : Multiply the coefficient of the first term by the constant   11 • -13 = -143 

Step-2 : Find two factors of  -143  whose sum equals the coefficient of the middle term, which is   -6 .

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

Equation at the end of step  2  :

  11n2 - 6n - 13  = 0 

Step  3  :

Parabola, Finding the Vertex :

 3.1      Find the Vertex of   y = 11n²-6n-13

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, 11 , 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,An²+Bn+C,the  n -coordinate of the vertex is given by  -B/(2A) . In our case the  n  coordinate is   0.2727  

 Plugging into the parabola formula   0.2727  for  n  we can calculate the  y -coordinate : 
  y = 11.0 * 0.27 * 0.27 - 6.0 * 0.27 - 13.0
or   y = -13.818

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 11n²-6n-13
Axis of Symmetry (dashed)  {n}={ 0.27} 
Vertex at  {n,y} = { 0.27,-13.82} 
 n -Intercepts (Roots) :
Root 1 at  {n,y} = {-0.85, 0.00} 
Root 2 at  {n,y} = { 1.39, 0.00} 

Solve Quadratic Equation by Completing The Square

 3.2     Solving   11n²-6n-13 = 0 by Completing The Square .

 Divide both sides of the equation by  11  to have 1 as the coefficient of the first term :
   n²-(6/11)n-(13/11) = 0

Add  13/11  to both side of the equation :
   n²-(6/11)n = 13/11

Now the clever bit: Take the coefficient of  n , which is  6/11 , divide by two, giving  3/11 , and finally square it giving  9/121 

Add  9/121  to both sides of the equation :
  On the right hand side we have :
   13/11  +  9/121   The common denominator of the two fractions is  121   Adding  (143/121)+(9/121)  gives  152/121 
  So adding to both sides we finally get :
   n²-(6/11)n+(9/121) = 152/121

Adding  9/121  has completed the left hand side into a perfect square :
   n²-(6/11)n+(9/121)  =
   (n-(3/11)) • (n-(3/11))  =
  (n-(3/11))²
Things which are equal to the same thing are also equal to one another. Since
   n²-(6/11)n+(9/121) = 152/121 and
   n²-(6/11)n+(9/121) = (n-(3/11))²
then, according to the law of transitivity,
   (n-(3/11))² = 152/121

We'll refer to this Equation as  Eq. #3.2.1  

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

Note that the square root of
   (n-(3/11))²   is
   (n-(3/11))^2/2 =
  (n-(3/11))¹ =
   n-(3/11)

Now, applying the Square Root Principle to  Eq. #3.2.1  we get:
   n-(3/11) = √ 152/121

Add  3/11  to both sides to obtain:
   n = 3/11 + √ 152/121

Since a square root has two values, one positive and the other negative
   n² - (6/11)n - (13/11) = 0
   has two solutions:
  n = 3/11 + √ 152/121
   or
  n = 3/11 - √ 152/121

Note that  √ 152/121 can be written as
  √ 152  / √ 121   which is √ 152  / 11

Solve Quadratic Equation using the Quadratic Formula

 3.3     Solving    11n²-6n-13 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  n  , the solution for   An²+Bn+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  n =   ————————
                      2A

  In our case,  A   =     11
                      B   =    -6
                      C   =  -13

Accordingly,  B²  -  4AC   =
                     36 - (-572) =
                     608

Applying the quadratic formula :

               6 ± √ 608
   n  =    —————
                    22

Can  √ 608 be simplified ?

Yes!   The prime factorization of  608   is
   2•2•2•2•2•19 
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).

√ 608   =  √ 2•2•2•2•2•19   =2•2•√ 38   =
                ±  4 • √ 38

  √ 38   , rounded to 4 decimal digits, is   6.1644
 So now we are looking at:
           n  =  ( 6 ± 4 •  6.164 ) / 22

Two real solutions:

 n =(6+√608)/22=(3+2√ 38 )/11= 1.394

or:

 n =(6-√608)/22=(3-2√ 38 )/11= -0.848

Two solutions were found :

1.  n =(6-√608)/22=(3-2√ 38 )/11= -0.848
2.  n =(6+√608)/22=(3+2√ 38 )/11= 1.394