Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((3•5x2) -  116x) +  220  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  15x²-116x+220 

The first term is,  15x²  its coefficient is  15 .
The middle term is,  -116x  its coefficient is  -116 .
The last term, "the constant", is  +220 

Step-1 : Multiply the coefficient of the first term by the constant   15 • 220 = 3300 

Step-2 : Find two factors of  3300  whose sum equals the coefficient of the middle term, which is   -116 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -66  and  -50 
                     15x² - 66x - 50x - 220

Step-4 : Add up the first 2 terms, pulling out like factors :
                    3x • (5x-22)
              Add up the last 2 terms, pulling out common factors :
                    10 • (5x-22)
Step-5 : Add up the four terms of step 4 :
                    (3x-10)  •  (5x-22)
             Which is the desired factorization

Equation at the end of step  2  :

  (5x - 22) • (3x - 10)  = 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  :    5x-22 = 0 

 Add  22  to both sides of the equation : 
                      5x = 22
Divide both sides of the equation by 5:
                     x = 22/5 = 4.400

Solving a Single Variable Equation :

 3.3      Solve  :    3x-10 = 0 

 Add  10  to both sides of the equation : 
                      3x = 10
Divide both sides of the equation by 3:
                     x = 10/3 = 3.333

Supplement : Solving Quadratic Equation Directly

Solving    15x2-116x+220  = 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 = 15x²-116x+220

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, 15 , 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   3.8667  

 Plugging into the parabola formula   3.8667  for  x  we can calculate the  y -coordinate : 
  y = 15.0 * 3.87 * 3.87 - 116.0 * 3.87 + 220.0
or   y = -4.267

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 15x²-116x+220
Axis of Symmetry (dashed)  {x}={ 3.87} 
Vertex at  {x,y} = { 3.87,-4.27} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = { 3.33, 0.00} 
Root 2 at  {x,y} = { 4.40, 0.00} 

Solve Quadratic Equation by Completing The Square

 4.2     Solving   15x²-116x+220 = 0 by Completing The Square .

 Divide both sides of the equation by  15  to have 1 as the coefficient of the first term :
   x²-(116/15)x+(44/3) = 0

Subtract  44/3  from both side of the equation :
   x²-(116/15)x = -44/3

Now the clever bit: Take the coefficient of  x , which is  116/15 , divide by two, giving  58/15 , and finally square it giving  3364/225 

Add  3364/225  to both sides of the equation :
  On the right hand side we have :
   -44/3  +  3364/225   The common denominator of the two fractions is  225   Adding  (-3300/225)+(3364/225)  gives  64/225 
  So adding to both sides we finally get :
   x²-(116/15)x+(3364/225) = 64/225

Adding  3364/225  has completed the left hand side into a perfect square :
   x²-(116/15)x+(3364/225)  =
   (x-(58/15)) • (x-(58/15))  =
  (x-(58/15))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(116/15)x+(3364/225) = 64/225 and
   x²-(116/15)x+(3364/225) = (x-(58/15))²
then, according to the law of transitivity,
   (x-(58/15))² = 64/225

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
   (x-(58/15))²   is
   (x-(58/15))^2/2 =
  (x-(58/15))¹ =
   x-(58/15)

Now, applying the Square Root Principle to  Eq. #4.2.1  we get:
   x-(58/15) = √ 64/225

Add  58/15  to both sides to obtain:
   x = 58/15 + √ 64/225

Since a square root has two values, one positive and the other negative
   x² - (116/15)x + (44/3) = 0
   has two solutions:
  x = 58/15 + √ 64/225
   or
  x = 58/15 - √ 64/225

Note that  √ 64/225 can be written as
  √ 64  / √ 225   which is 8 / 15

Solve Quadratic Equation using the Quadratic Formula

 4.3     Solving    15x²-116x+220 = 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   =     15
                      B   =   -116
                      C   =  220

Accordingly,  B²  -  4AC   =
                     13456 - 13200 =
                     256

Applying the quadratic formula :

               116 ± √ 256
   x  =    ——————
                      30

Can  √ 256 be simplified ?

Yes!   The prime factorization of  256   is
   2•2•2•2•2•2•2•2 
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).

√ 256   =  √ 2•2•2•2•2•2•2•2   =2•2•2•2•√ 1   =
                ±  16 • √ 1   =
                ±  16

So now we are looking at:
           x  =  ( 116 ± 16) / 30

Two real solutions:

x =(116+√256)/30=(58+8)/15= 4.400

or:

x =(116-√256)/30=(58-8)/15= 3.333

Two solutions were found :

1.  x = 10/3 = 3.333
   
2.  x = 22/5 = 4.400