Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((0 -  (3•5x2)) -  11x) -  2  = 0 

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   -15x² - 11x - 2  =   -1 • (15x² + 11x + 2) 

Trying to factor by splitting the middle term

 3.2     Factoring  15x² + 11x + 2 

The first term is,  15x²  its coefficient is  15 .
The middle term is,  +11x  its coefficient is  11 .
The last term, "the constant", is  +2 

Step-1 : Multiply the coefficient of the first term by the constant   15 • 2 = 30 

Step-2 : Find two factors of  30  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,  5  and  6 
                     15x² + 5x + 6x + 2

Step-4 : Add up the first 2 terms, pulling out like factors :
                    5x • (3x+1)
              Add up the last 2 terms, pulling out common factors :
                    2 • (3x+1)
Step-5 : Add up the four terms of step 4 :
                    (5x+2)  •  (3x+1)
             Which is the desired factorization

Equation at the end of step  3  :

  (-3x - 1) • (5x + 2)  = 0 

Step  4  :

Theory - Roots of a product :

 4.1    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 :

 4.2      Solve  :    -3x-1 = 0 

 Add  1  to both sides of the equation : 
                      -3x = 1
Multiply both sides of the equation by (-1) :  3x = -1


Divide both sides of the equation by 3:
                     x = -1/3 = -0.333

Solving a Single Variable Equation :

 4.3      Solve  :    5x+2 = 0 

 Subtract  2  from both sides of the equation : 
                      5x = -2
Divide both sides of the equation by 5:
                     x = -2/5 = -0.400

Supplement : Solving Quadratic Equation Directly

Solving    15x2+11x+2  = 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 :

 5.1      Find the Vertex of   y = 15x²+11x+2

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  -0.3667  

 Plugging into the parabola formula  -0.3667  for  x  we can calculate the  y -coordinate : 
  y = 15.0 * -0.37 * -0.37 + 11.0 * -0.37 + 2.0
or   y = -0.017

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 15x²+11x+2
Axis of Symmetry (dashed)  {x}={-0.37} 
Vertex at  {x,y} = {-0.37,-0.02} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-0.40, 0.00} 
Root 2 at  {x,y} = {-0.33, 0.00} 

Solve Quadratic Equation by Completing The Square

 5.2     Solving   15x²+11x+2 = 0 by Completing The Square .

 Divide both sides of the equation by  15  to have 1 as the coefficient of the first term :
   x²+(11/15)x+(2/15) = 0

Subtract  2/15  from both side of the equation :
   x²+(11/15)x = -2/15

Now the clever bit: Take the coefficient of  x , which is  11/15 , divide by two, giving  11/30 , and finally square it giving  121/900 

Add  121/900  to both sides of the equation :
  On the right hand side we have :
   -2/15  +  121/900   The common denominator of the two fractions is  900   Adding  (-120/900)+(121/900)  gives  1/900 
  So adding to both sides we finally get :
   x²+(11/15)x+(121/900) = 1/900

Adding  121/900  has completed the left hand side into a perfect square :
   x²+(11/15)x+(121/900)  =
   (x+(11/30)) • (x+(11/30))  =
  (x+(11/30))²
Things which are equal to the same thing are also equal to one another. Since
   x²+(11/15)x+(121/900) = 1/900 and
   x²+(11/15)x+(121/900) = (x+(11/30))²
then, according to the law of transitivity,
   (x+(11/30))² = 1/900

We'll refer to this Equation as  Eq. #5.2.1  

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
   (x+(11/30))²   is
   (x+(11/30))^2/2 =
  (x+(11/30))¹ =
   x+(11/30)

Now, applying the Square Root Principle to  Eq. #5.2.1  we get:
   x+(11/30) = √ 1/900

Subtract  11/30  from both sides to obtain:
   x = -11/30 + √ 1/900

Since a square root has two values, one positive and the other negative
   x² + (11/15)x + (2/15) = 0
   has two solutions:
  x = -11/30 + √ 1/900
   or
  x = -11/30 - √ 1/900

Note that  √ 1/900 can be written as
  √ 1  / √ 900   which is 1 / 30

Solve Quadratic Equation using the Quadratic Formula

 5.3     Solving    15x²+11x+2 = 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   =    11
                      C   =   2

Accordingly,  B²  -  4AC   =
                     121 - 120 =
                     1

Applying the quadratic formula :

               -11 ± √ 1
   x  =    —————
                    30

So now we are looking at:
           x  =  ( -11 ± 1) / 30

Two real solutions:

x =(-11+√1)/30=-0.333

or:

x =(-11-√1)/30=-0.400

Two solutions were found :

1.  x = -2/5 = -0.400
   
2.  x = -1/3 = -0.333