Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((2•3•5x2) -  61x) +  30  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  30x²-61x+30 

The first term is,  30x²  its coefficient is  30 .
The middle term is,  -61x  its coefficient is  -61 .
The last term, "the constant", is  +30 

Step-1 : Multiply the coefficient of the first term by the constant   30 • 30 = 900 

Step-2 : Find two factors of  900  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,  -36  and  -25 
                     30x² - 36x - 25x - 30

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

Equation at the end of step  2  :

  (5x - 6) • (6x - 5)  = 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-6 = 0 

 Add  6  to both sides of the equation : 
                      5x = 6
Divide both sides of the equation by 5:
                     x = 6/5 = 1.200

Solving a Single Variable Equation :

 3.3      Solve  :    6x-5 = 0 

 Add  5  to both sides of the equation : 
                      6x = 5
Divide both sides of the equation by 6:
                     x = 5/6 = 0.833

Supplement : Solving Quadratic Equation Directly

Solving    30x2-61x+30  = 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 = 30x²-61x+30

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, 30 , 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   1.0167  

 Plugging into the parabola formula   1.0167  for  x  we can calculate the  y -coordinate : 
  y = 30.0 * 1.02 * 1.02 - 61.0 * 1.02 + 30.0
or   y = -1.008

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 30x²-61x+30
Axis of Symmetry (dashed)  {x}={ 1.02} 
Vertex at  {x,y} = { 1.02,-1.01} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = { 0.83, 0.00} 
Root 2 at  {x,y} = { 1.20, 0.00} 

Solve Quadratic Equation by Completing The Square

 4.2     Solving   30x²-61x+30 = 0 by Completing The Square .

 Divide both sides of the equation by  30  to have 1 as the coefficient of the first term :
   x²-(61/30)x+1 = 0

Subtract  1  from both side of the equation :
   x²-(61/30)x = -1

Now the clever bit: Take the coefficient of  x , which is  61/30 , divide by two, giving  61/60 , and finally square it giving  3721/3600 

Add  3721/3600  to both sides of the equation :
  On the right hand side we have :
   -1  +  3721/3600    or,  (-1/1)+(3721/3600) 
  The common denominator of the two fractions is  3600   Adding  (-3600/3600)+(3721/3600)  gives  121/3600 
  So adding to both sides we finally get :
   x²-(61/30)x+(3721/3600) = 121/3600

Adding  3721/3600  has completed the left hand side into a perfect square :
   x²-(61/30)x+(3721/3600)  =
   (x-(61/60)) • (x-(61/60))  =
  (x-(61/60))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(61/30)x+(3721/3600) = 121/3600 and
   x²-(61/30)x+(3721/3600) = (x-(61/60))²
then, according to the law of transitivity,
   (x-(61/60))² = 121/3600

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-(61/60))²   is
   (x-(61/60))^2/2 =
  (x-(61/60))¹ =
   x-(61/60)

Now, applying the Square Root Principle to  Eq. #4.2.1  we get:
   x-(61/60) = √ 121/3600

Add  61/60  to both sides to obtain:
   x = 61/60 + √ 121/3600

Since a square root has two values, one positive and the other negative
   x² - (61/30)x + 1 = 0
   has two solutions:
  x = 61/60 + √ 121/3600
   or
  x = 61/60 - √ 121/3600

Note that  √ 121/3600 can be written as
  √ 121  / √ 3600   which is 11 / 60

Solve Quadratic Equation using the Quadratic Formula

 4.3     Solving    30x²-61x+30 = 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   =     30
                      B   =   -61
                      C   =   30

Accordingly,  B²  -  4AC   =
                     3721 - 3600 =
                     121

Applying the quadratic formula :

               61 ± √ 121
   x  =    ——————
                      60

Can  √ 121 be simplified ?

Yes!   The prime factorization of  121   is
   11•11 
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).

√ 121   =  √ 11•11   =
                ±  11 • √ 1   =
                ±  11

So now we are looking at:
           x  =  ( 61 ± 11) / 60

Two real solutions:

x =(61+√121)/60=(61+11)/60= 1.200

or:

x =(61-√121)/60=(61-11)/60= 0.833

Two solutions were found :

1.  x = 5/6 = 0.833
   
2.  x = 6/5 = 1.200