Solution - Quadratic equations

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                     63*z^2+61*z-(-6)=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((32•7z2) +  61z) -  -6  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  63z²+61z+6 

The first term is,  63z²  its coefficient is  63 .
The middle term is,  +61z  its coefficient is  61 .
The last term, "the constant", is  +6 

Step-1 : Multiply the coefficient of the first term by the constant   63 • 6 = 378 

Step-2 : Find two factors of  378  whose sum equals the coefficient of the middle term, which is   61 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  7  and  54 
                     63z² + 7z + 54z + 6

Step-4 : Add up the first 2 terms, pulling out like factors :
                    7z • (9z+1)
              Add up the last 2 terms, pulling out common factors :
                    6 • (9z+1)
Step-5 : Add up the four terms of step 4 :
                    (7z+6)  •  (9z+1)
             Which is the desired factorization

Equation at the end of step  2  :

  (9z + 1) • (7z + 6)  = 0 

Step  3  :

Theory - Roots of a product :

 3.1    A product of several terms equals zero. 

 When a product of two or more terms equals zero, then at least one of the terms must be zero. 

 We shall now solve each term = 0 separately 

 In other words, we are going to solve as many equations as there are terms in the product 

 Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

 3.2      Solve  :    9z+1 = 0 

 Subtract  1  from both sides of the equation : 
                      9z = -1
Divide both sides of the equation by 9:
                     z = -1/9 = -0.111

Solving a Single Variable Equation :

 3.3      Solve  :    7z+6 = 0 

 Subtract  6  from both sides of the equation : 
                      7z = -6
Divide both sides of the equation by 7:
                     z = -6/7 = -0.857

Supplement : Solving Quadratic Equation Directly

Solving    63z2+61z+6  = 0   directly 

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex :

 4.1      Find the Vertex of   y = 63z²+61z+6

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, 63 , 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  -0.4841  

 Plugging into the parabola formula  -0.4841  for  z  we can calculate the  y -coordinate : 
  y = 63.0 * -0.48 * -0.48 + 61.0 * -0.48 + 6.0
or   y = -8.766

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 63z²+61z+6
Axis of Symmetry (dashed)  {z}={-0.48} 
Vertex at  {z,y} = {-0.48,-8.77} 
 z -Intercepts (Roots) :
Root 1 at  {z,y} = {-0.86, 0.00} 
Root 2 at  {z,y} = {-0.11, 0.00} 

Solve Quadratic Equation by Completing The Square

 4.2     Solving   63z²+61z+6 = 0 by Completing The Square .

 Divide both sides of the equation by  63  to have 1 as the coefficient of the first term :
   z²+(61/63)z+(2/21) = 0

Subtract  2/21  from both side of the equation :
   z²+(61/63)z = -2/21

Now the clever bit: Take the coefficient of  z , which is  61/63 , divide by two, giving  61/126 , and finally square it giving  3721/15876 

Add  3721/15876  to both sides of the equation :
  On the right hand side we have :
   -2/21  +  3721/15876   The common denominator of the two fractions is  15876   Adding  (-1512/15876)+(3721/15876)  gives  2209/15876 
  So adding to both sides we finally get :
   z²+(61/63)z+(3721/15876) = 2209/15876

Adding  3721/15876  has completed the left hand side into a perfect square :
   z²+(61/63)z+(3721/15876)  =
   (z+(61/126)) • (z+(61/126))  =
  (z+(61/126))²
Things which are equal to the same thing are also equal to one another. Since
   z²+(61/63)z+(3721/15876) = 2209/15876 and
   z²+(61/63)z+(3721/15876) = (z+(61/126))²
then, according to the law of transitivity,
   (z+(61/126))² = 2209/15876

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
   (z+(61/126))²   is
   (z+(61/126))^2/2 =
  (z+(61/126))¹ =
   z+(61/126)

Now, applying the Square Root Principle to  Eq. #4.2.1  we get:
   z+(61/126) = √ 2209/15876

Subtract  61/126  from both sides to obtain:
   z = -61/126 + √ 2209/15876

Since a square root has two values, one positive and the other negative
   z² + (61/63)z + (2/21) = 0
   has two solutions:
  z = -61/126 + √ 2209/15876
   or
  z = -61/126 - √ 2209/15876

Note that  √ 2209/15876 can be written as
  √ 2209  / √ 15876   which is 47 / 126

Solve Quadratic Equation using the Quadratic Formula

 4.3     Solving    63z²+61z+6 = 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   =     63
                      B   =    61
                      C   =   6

Accordingly,  B²  -  4AC   =
                     3721 - 1512 =
                     2209

Applying the quadratic formula :

               -61 ± √ 2209
   z  =    ———————
                        126

Can  √ 2209 be simplified ?

Yes!   The prime factorization of  2209   is
   47•47 
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).

√ 2209   =  √ 47•47   =
                ±  47 • √ 1   =
                ±  47

So now we are looking at:
           z  =  ( -61 ± 47) / 126

Two real solutions:

z =(-61+√2209)/126=(-61+47)/126= -0.111

or:

z =(-61-√2209)/126=(-61-47)/126= -0.857

Two solutions were found :

1.  z = -6/7 = -0.857
   
2.  z = -1/9 = -0.111