Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((23•5x2) -  131x) +  105  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  40x²-131x+105 

The first term is,  40x²  its coefficient is  40 .
The middle term is,  -131x  its coefficient is  -131 .
The last term, "the constant", is  +105 

Step-1 : Multiply the coefficient of the first term by the constant   40 • 105 = 4200 

Step-2 : Find two factors of  4200  whose sum equals the coefficient of the middle term, which is   -131 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -75  and  -56 
                     40x² - 75x - 56x - 105

Step-4 : Add up the first 2 terms, pulling out like factors :
                    5x • (8x-15)
              Add up the last 2 terms, pulling out common factors :
                    7 • (8x-15)
Step-5 : Add up the four terms of step 4 :
                    (5x-7)  •  (8x-15)
             Which is the desired factorization

Equation at the end of step  2  :

  (8x - 15) • (5x - 7)  = 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  :    8x-15 = 0 

 Add  15  to both sides of the equation : 
                      8x = 15
Divide both sides of the equation by 8:
                     x = 15/8 = 1.875

Solving a Single Variable Equation :

 3.3      Solve  :    5x-7 = 0 

 Add  7  to both sides of the equation : 
                      5x = 7
Divide both sides of the equation by 5:
                     x = 7/5 = 1.400

Supplement : Solving Quadratic Equation Directly

Solving    40x2-131x+105  = 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 = 40x²-131x+105

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, 40 , 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   1.6375  

 Plugging into the parabola formula   1.6375  for  x  we can calculate the  y -coordinate : 
  y = 40.0 * 1.64 * 1.64 - 131.0 * 1.64 + 105.0
or   y = -2.256

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 40x²-131x+105
Axis of Symmetry (dashed)  {x}={ 1.64} 
Vertex at  {x,y} = { 1.64,-2.26} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = { 1.40, 0.00} 
Root 2 at  {x,y} = { 1.88, 0.00} 

Solve Quadratic Equation by Completing The Square

 4.2     Solving   40x²-131x+105 = 0 by Completing The Square .

 Divide both sides of the equation by  40  to have 1 as the coefficient of the first term :
   x²-(131/40)x+(21/8) = 0

Subtract  21/8  from both side of the equation :
   x²-(131/40)x = -21/8

Now the clever bit: Take the coefficient of  x , which is  131/40 , divide by two, giving  131/80 , and finally square it giving  17161/6400 

Add  17161/6400  to both sides of the equation :
  On the right hand side we have :
   -21/8  +  17161/6400   The common denominator of the two fractions is  6400   Adding  (-16800/6400)+(17161/6400)  gives  361/6400 
  So adding to both sides we finally get :
   x²-(131/40)x+(17161/6400) = 361/6400

Adding  17161/6400  has completed the left hand side into a perfect square :
   x²-(131/40)x+(17161/6400)  =
   (x-(131/80)) • (x-(131/80))  =
  (x-(131/80))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(131/40)x+(17161/6400) = 361/6400 and
   x²-(131/40)x+(17161/6400) = (x-(131/80))²
then, according to the law of transitivity,
   (x-(131/80))² = 361/6400

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-(131/80))²   is
   (x-(131/80))^2/2 =
  (x-(131/80))¹ =
   x-(131/80)

Now, applying the Square Root Principle to  Eq. #4.2.1  we get:
   x-(131/80) = √ 361/6400

Add  131/80  to both sides to obtain:
   x = 131/80 + √ 361/6400

Since a square root has two values, one positive and the other negative
   x² - (131/40)x + (21/8) = 0
   has two solutions:
  x = 131/80 + √ 361/6400
   or
  x = 131/80 - √ 361/6400

Note that  √ 361/6400 can be written as
  √ 361  / √ 6400   which is 19 / 80

Solve Quadratic Equation using the Quadratic Formula

 4.3     Solving    40x²-131x+105 = 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   =     40
                      B   =   -131
                      C   =  105

Accordingly,  B²  -  4AC   =
                     17161 - 16800 =
                     361

Applying the quadratic formula :

               131 ± √ 361
   x  =    ——————
                      80

Can  √ 361 be simplified ?

Yes!   The prime factorization of  361   is
   19•19 
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).

√ 361   =  √ 19•19   =
                ±  19 • √ 1   =
                ±  19

So now we are looking at:
           x  =  ( 131 ± 19) / 80

Two real solutions:

x =(131+√361)/80=(131+19)/80= 1.875

or:

x =(131-√361)/80=(131-19)/80= 1.400

Two solutions were found :

1.  x = 7/5 = 1.400
   
2.  x = 15/8 = 1.875