Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (32x2 -  11x) -  39  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  9x²-11x-39 

The first term is,  9x²  its coefficient is  9 .
The middle term is,  -11x  its coefficient is  -11 .
The last term, "the constant", is  -39 

Step-1 : Multiply the coefficient of the first term by the constant   9 • -39 = -351 

Step-2 : Find two factors of  -351  whose sum equals the coefficient of the middle term, which is   -11 .

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

Equation at the end of step  2  :

  9x2 - 11x - 39  = 0 

Step  3  :

Parabola, Finding the Vertex :

 3.1      Find the Vertex of   y = 9x²-11x-39

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, 9 , 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   0.6111  

 Plugging into the parabola formula   0.6111  for  x  we can calculate the  y -coordinate : 
  y = 9.0 * 0.61 * 0.61 - 11.0 * 0.61 - 39.0
or   y = -42.361

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 9x²-11x-39
Axis of Symmetry (dashed)  {x}={ 0.61} 
Vertex at  {x,y} = { 0.61,-42.36} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-1.56, 0.00} 
Root 2 at  {x,y} = { 2.78, 0.00} 

Solve Quadratic Equation by Completing The Square

 3.2     Solving   9x²-11x-39 = 0 by Completing The Square .

 Divide both sides of the equation by  9  to have 1 as the coefficient of the first term :
   x²-(11/9)x-(13/3) = 0

Add  13/3  to both side of the equation :
   x²-(11/9)x = 13/3

Now the clever bit: Take the coefficient of  x , which is  11/9 , divide by two, giving  11/18 , and finally square it giving  121/324 

Add  121/324  to both sides of the equation :
  On the right hand side we have :
   13/3  +  121/324   The common denominator of the two fractions is  324   Adding  (1404/324)+(121/324)  gives  1525/324 
  So adding to both sides we finally get :
   x²-(11/9)x+(121/324) = 1525/324

Adding  121/324  has completed the left hand side into a perfect square :
   x²-(11/9)x+(121/324)  =
   (x-(11/18)) • (x-(11/18))  =
  (x-(11/18))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(11/9)x+(121/324) = 1525/324 and
   x²-(11/9)x+(121/324) = (x-(11/18))²
then, according to the law of transitivity,
   (x-(11/18))² = 1525/324

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
   (x-(11/18))²   is
   (x-(11/18))^2/2 =
  (x-(11/18))¹ =
   x-(11/18)

Now, applying the Square Root Principle to  Eq. #3.2.1  we get:
   x-(11/18) = √ 1525/324

Add  11/18  to both sides to obtain:
   x = 11/18 + √ 1525/324

Since a square root has two values, one positive and the other negative
   x² - (11/9)x - (13/3) = 0
   has two solutions:
  x = 11/18 + √ 1525/324
   or
  x = 11/18 - √ 1525/324

Note that  √ 1525/324 can be written as
  √ 1525  / √ 324   which is √ 1525  / 18

Solve Quadratic Equation using the Quadratic Formula

 3.3     Solving    9x²-11x-39 = 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   =     9
                      B   =   -11
                      C   =  -39

Accordingly,  B²  -  4AC   =
                     121 - (-1404) =
                     1525

Applying the quadratic formula :

               11 ± √ 1525
   x  =    ——————
                      18

Can  √ 1525 be simplified ?

Yes!   The prime factorization of  1525   is
   5•5•61 
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).

√ 1525   =  √ 5•5•61   =
                ±  5 • √ 61

  √ 61   , rounded to 4 decimal digits, is   7.8102
 So now we are looking at:
           x  =  ( 11 ± 5 •  7.810 ) / 18

Two real solutions:

 x =(11+√1525)/18=(11+5√ 61 )/18= 2.781

or:

 x =(11-√1525)/18=(11-5√ 61 )/18= -1.558

Two solutions were found :

1.  x =(11-√1525)/18=(11-5√ 61 )/18= -1.558
2.  x =(11+√1525)/18=(11+5√ 61 )/18= 2.781