Solution - Quadratic equations

Step by step solution :

Step  1  :

Trying to factor by splitting the middle term

 1.1     Factoring  v²-28v-1 

The first term is,  v²  its coefficient is  1 .
The middle term is,  -28v  its coefficient is  -28 .
The last term, "the constant", is  -1 

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

Step-2 : Find two factors of  -1  whose sum equals the coefficient of the middle term, which is   -28 .

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Equation at the end of step  1  :

  v2 - 28v - 1  = 0 

Step  2  :

Parabola, Finding the Vertex :

 2.1      Find the Vertex of   y = v²-28v-1

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, 1 , 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,Av²+Bv+C,the  v -coordinate of the vertex is given by  -B/(2A) . In our case the  v  coordinate is  14.0000  

 Plugging into the parabola formula  14.0000  for  v  we can calculate the  y -coordinate : 
  y = 1.0 * 14.00 * 14.00 - 28.0 * 14.00 - 1.0
or   y = -197.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = v²-28v-1
Axis of Symmetry (dashed)  {v}={14.00} 
Vertex at  {v,y} = {14.00,-197.00} 
 v -Intercepts (Roots) :
Root 1 at  {v,y} = {-0.04, 0.00} 
Root 2 at  {v,y} = {28.04, 0.00} 

Solve Quadratic Equation by Completing The Square

 2.2     Solving   v²-28v-1 = 0 by Completing The Square .

 Add  1  to both side of the equation :
   v²-28v = 1

Now the clever bit: Take the coefficient of  v , which is  28 , divide by two, giving  14 , and finally square it giving  196 

Add  196  to both sides of the equation :
  On the right hand side we have :
   1  +  196    or,  (1/1)+(196/1) 
  The common denominator of the two fractions is  1   Adding  (1/1)+(196/1)  gives  197/1 
  So adding to both sides we finally get :
   v²-28v+196 = 197

Adding  196  has completed the left hand side into a perfect square :
   v²-28v+196  =
   (v-14) • (v-14)  =
  (v-14)²
Things which are equal to the same thing are also equal to one another. Since
   v²-28v+196 = 197 and
   v²-28v+196 = (v-14)²
then, according to the law of transitivity,
   (v-14)² = 197

We'll refer to this Equation as  Eq. #2.2.1  

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

Note that the square root of
   (v-14)²   is
   (v-14)^2/2 =
  (v-14)¹ =
   v-14

Now, applying the Square Root Principle to  Eq. #2.2.1  we get:
   v-14 = √ 197

Add  14  to both sides to obtain:
   v = 14 + √ 197

Since a square root has two values, one positive and the other negative
   v² - 28v - 1 = 0
   has two solutions:
  v = 14 + √ 197
   or
  v = 14 - √ 197

Solve Quadratic Equation using the Quadratic Formula

 2.3     Solving    v²-28v-1 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  v  , the solution for   Av²+Bv+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  v =   ————————
                      2A

  In our case,  A   =     1
                      B   =   -28
                      C   =   -1

Accordingly,  B²  -  4AC   =
                     784 - (-4) =
                     788

Applying the quadratic formula :

               28 ± √ 788
   v  =    ——————
                      2

Can  √ 788 be simplified ?

Yes!   The prime factorization of  788   is
   2•2•197 
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).

√ 788   =  √ 2•2•197   =
                ±  2 • √ 197

  √ 197   , rounded to 4 decimal digits, is  14.0357
 So now we are looking at:
           v  =  ( 28 ± 2 •  14.036 ) / 2

Two real solutions:

 v =(28+√788)/2=14+√ 197 = 28.036

or:

 v =(28-√788)/2=14-√ 197 = -0.036

Two solutions were found :

1.  v =(28-√788)/2=14-√ 197 = -0.036
2.  v =(28+√788)/2=14+√ 197 = 28.036