Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((2•3z2) +  z) +  10  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  6z²+z+10 

The first term is,  6z²  its coefficient is  6 .
The middle term is,  +z  its coefficient is  1 .
The last term, "the constant", is  +10 

Step-1 : Multiply the coefficient of the first term by the constant   6 • 10 = 60 

Step-2 : Find two factors of  60  whose sum equals the coefficient of the middle term, which is   1 .

For tidiness, printing of 18 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  2  :

  6z2 + z + 10  = 0 

Step  3  :

Parabola, Finding the Vertex :

 3.1      Find the Vertex of   y = 6z²+z+10

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, 6 , 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,Az²+Bz+C,the  z -coordinate of the vertex is given by  -B/(2A) . In our case the  z  coordinate is  -0.0833  

 Plugging into the parabola formula  -0.0833  for  z  we can calculate the  y -coordinate : 
  y = 6.0 * -0.08 * -0.08 + 1.0 * -0.08 + 10.0
or   y = 9.958

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 6z²+z+10
Axis of Symmetry (dashed)  {z}={-0.08} 
Vertex at  {z,y} = {-0.08, 9.96} 
Function has no real roots

Solve Quadratic Equation by Completing The Square

 3.2     Solving   6z²+z+10 = 0 by Completing The Square .

 Divide both sides of the equation by  6  to have 1 as the coefficient of the first term :
   z²+(1/6)z+(5/3) = 0

Subtract  5/3  from both side of the equation :
   z²+(1/6)z = -5/3

Now the clever bit: Take the coefficient of  z , which is  1/6 , divide by two, giving  1/12 , and finally square it giving  1/144 

Add  1/144  to both sides of the equation :
  On the right hand side we have :
   -5/3  +  1/144   The common denominator of the two fractions is  144   Adding  (-240/144)+(1/144)  gives  -239/144 
  So adding to both sides we finally get :
   z²+(1/6)z+(1/144) = -239/144

Adding  1/144  has completed the left hand side into a perfect square :
   z²+(1/6)z+(1/144)  =
   (z+(1/12)) • (z+(1/12))  =
  (z+(1/12))²
Things which are equal to the same thing are also equal to one another. Since
   z²+(1/6)z+(1/144) = -239/144 and
   z²+(1/6)z+(1/144) = (z+(1/12))²
then, according to the law of transitivity,
   (z+(1/12))² = -239/144

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
   (z+(1/12))²   is
   (z+(1/12))^2/2 =
  (z+(1/12))¹ =
   z+(1/12)

Now, applying the Square Root Principle to  Eq. #3.2.1  we get:
   z+(1/12) = √ -239/144

Subtract  1/12  from both sides to obtain:
   z = -1/12 + √ -239/144
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
   z² + (1/6)z + (5/3) = 0
   has two solutions:
  z = -1/12 + √ 239/144 •  i 
   or
  z = -1/12 - √ 239/144 •  i 

Note that  √ 239/144 can be written as
  √ 239  / √ 144   which is √ 239  / 12

Solve Quadratic Equation using the Quadratic Formula

 3.3     Solving    6z²+z+10 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  z  , the solution for   Az²+Bz+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  z =   ————————
                      2A

  In our case,  A   =     6
                      B   =    1
                      C   =   10

Accordingly,  B²  -  4AC   =
                     1 - 240 =
                     -239

Applying the quadratic formula :

               -1 ± √ -239
   z  =    ——————
                      12

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,√ -239  = 
                    √ 239 • (-1)  =
                    √ 239  • √ -1   =
                    ±  √ 239  • i

  √ 239   , rounded to 4 decimal digits, is  15.4596
 So now we are looking at:
           z  =  ( -1 ±  15.460 i ) / 12

Two imaginary solutions :

 z =(-1+√-239)/12=(-1+i√ 239 )/12= -0.0833+1.2883i
  or: 
 z =(-1-√-239)/12=(-1-i√ 239 )/12= -0.0833-1.2883i

Two solutions were found :

1.  z =(-1-√-239)/12=(-1-i√ 239 )/12= -0.0833-1.2883i
2.  z =(-1+√-239)/12=(-1+i√ 239 )/12= -0.0833+1.2883i