Solution - Quadratic equations

Rearrange:

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

                     2*c^2-9*c+5-(2*c)=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((2c2 -  9c) +  5) -  2c  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  2c²-11c+5 

The first term is,  2c²  its coefficient is  2 .
The middle term is,  -11c  its coefficient is  -11 .
The last term, "the constant", is  +5 

Step-1 : Multiply the coefficient of the first term by the constant   2 • 5 = 10 

Step-2 : Find two factors of  10  whose sum equals the coefficient of the middle term, which is   -11 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -10  and  -1 
                     2c² - 10c - 1c - 5

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

Equation at the end of step  2  :

  (c - 5) • (2c - 1)  = 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  :    c-5 = 0 

 Add  5  to both sides of the equation : 
                      c = 5

Solving a Single Variable Equation :

 3.3      Solve  :    2c-1 = 0 

 Add  1  to both sides of the equation : 
                      2c = 1
Divide both sides of the equation by 2:
                     c = 1/2 = 0.500

Supplement : Solving Quadratic Equation Directly

Solving    2c2-11c+5  = 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 = 2c²-11c+5

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, 2 , 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,Ac²+Bc+C,the  c -coordinate of the vertex is given by  -B/(2A) . In our case the  c  coordinate is   2.7500  

 Plugging into the parabola formula   2.7500  for  c  we can calculate the  y -coordinate : 
  y = 2.0 * 2.75 * 2.75 - 11.0 * 2.75 + 5.0
or   y = -10.125

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 2c²-11c+5
Axis of Symmetry (dashed)  {c}={ 2.75} 
Vertex at  {c,y} = { 2.75,-10.12} 
 c -Intercepts (Roots) :
Root 1 at  {c,y} = { 0.50, 0.00} 
Root 2 at  {c,y} = { 5.00, 0.00} 

Solve Quadratic Equation by Completing The Square

 4.2     Solving   2c²-11c+5 = 0 by Completing The Square .

 Divide both sides of the equation by  2  to have 1 as the coefficient of the first term :
   c²-(11/2)c+(5/2) = 0

Subtract  5/2  from both side of the equation :
   c²-(11/2)c = -5/2

Now the clever bit: Take the coefficient of  c , which is  11/2 , divide by two, giving  11/4 , and finally square it giving  121/16 

Add  121/16  to both sides of the equation :
  On the right hand side we have :
   -5/2  +  121/16   The common denominator of the two fractions is  16   Adding  (-40/16)+(121/16)  gives  81/16 
  So adding to both sides we finally get :
   c²-(11/2)c+(121/16) = 81/16

Adding  121/16  has completed the left hand side into a perfect square :
   c²-(11/2)c+(121/16)  =
   (c-(11/4)) • (c-(11/4))  =
  (c-(11/4))²
Things which are equal to the same thing are also equal to one another. Since
   c²-(11/2)c+(121/16) = 81/16 and
   c²-(11/2)c+(121/16) = (c-(11/4))²
then, according to the law of transitivity,
   (c-(11/4))² = 81/16

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
   (c-(11/4))²   is
   (c-(11/4))^2/2 =
  (c-(11/4))¹ =
   c-(11/4)

Now, applying the Square Root Principle to  Eq. #4.2.1  we get:
   c-(11/4) = √ 81/16

Add  11/4  to both sides to obtain:
   c = 11/4 + √ 81/16

Since a square root has two values, one positive and the other negative
   c² - (11/2)c + (5/2) = 0
   has two solutions:
  c = 11/4 + √ 81/16
   or
  c = 11/4 - √ 81/16

Note that  √ 81/16 can be written as
  √ 81  / √ 16   which is 9 / 4

Solve Quadratic Equation using the Quadratic Formula

 4.3     Solving    2c²-11c+5 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  c  , the solution for   Ac²+Bc+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  c =   ————————
                      2A

  In our case,  A   =     2
                      B   =   -11
                      C   =   5

Accordingly,  B²  -  4AC   =
                     121 - 40 =
                     81

Applying the quadratic formula :

               11 ± √ 81
   c  =    —————
                    4

Can  √ 81 be simplified ?

Yes!   The prime factorization of  81   is
   3•3•3•3 
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).

√ 81   =  √ 3•3•3•3   =3•3•√ 1   =
                ±  9 • √ 1   =
                ±  9

So now we are looking at:
           c  =  ( 11 ± 9) / 4

Two real solutions:

c =(11+√81)/4=(11+9)/4= 5.000

or:

c =(11-√81)/4=(11-9)/4= 0.500

Two solutions were found :

1.  c = 1/2 = 0.500
   
2.  c = 5