Solution - Quadratic equations

Step by step solution :

Step  1  :

Trying to factor by splitting the middle term

 1.1     Factoring  n²+2n-1095 

The first term is,  n²  its coefficient is  1 .
The middle term is,  +2n  its coefficient is  2 .
The last term, "the constant", is  -1095 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -1095 = -1095 

Step-2 : Find two factors of  -1095  whose sum equals the coefficient of the middle term, which is   2 .

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

Equation at the end of step  1  :

  n2 + 2n - 1095  = 0 

Step  2  :

Parabola, Finding the Vertex :

 2.1      Find the Vertex of   y = n²+2n-1095

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,An²+Bn+C,the  n -coordinate of the vertex is given by  -B/(2A) . In our case the  n  coordinate is  -1.0000  

 Plugging into the parabola formula  -1.0000  for  n  we can calculate the  y -coordinate : 
  y = 1.0 * -1.00 * -1.00 + 2.0 * -1.00 - 1095.0
or   y = -1096.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = n²+2n-1095
Axis of Symmetry (dashed)  {n}={-1.00} 
Vertex at  {n,y} = {-1.00,-1096.00} 
 n -Intercepts (Roots) :
Root 1 at  {n,y} = {-34.11, 0.00} 
Root 2 at  {n,y} = {32.11, 0.00} 

Solve Quadratic Equation by Completing The Square

 2.2     Solving   n²+2n-1095 = 0 by Completing The Square .

 Add  1095  to both side of the equation :
   n²+2n = 1095

Now the clever bit: Take the coefficient of  n , which is  2 , divide by two, giving  1 , and finally square it giving  1 

Add  1  to both sides of the equation :
  On the right hand side we have :
   1095  +  1    or,  (1095/1)+(1/1) 
  The common denominator of the two fractions is  1   Adding  (1095/1)+(1/1)  gives  1096/1 
  So adding to both sides we finally get :
   n²+2n+1 = 1096

Adding  1  has completed the left hand side into a perfect square :
   n²+2n+1  =
   (n+1) • (n+1)  =
  (n+1)²
Things which are equal to the same thing are also equal to one another. Since
   n²+2n+1 = 1096 and
   n²+2n+1 = (n+1)²
then, according to the law of transitivity,
   (n+1)² = 1096

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
   (n+1)²   is
   (n+1)^2/2 =
  (n+1)¹ =
   n+1

Now, applying the Square Root Principle to  Eq. #2.2.1  we get:
   n+1 = √ 1096

Subtract  1  from both sides to obtain:
   n = -1 + √ 1096

Since a square root has two values, one positive and the other negative
   n² + 2n - 1095 = 0
   has two solutions:
  n = -1 + √ 1096
   or
  n = -1 - √ 1096

Solve Quadratic Equation using the Quadratic Formula

 2.3     Solving    n²+2n-1095 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  n  , the solution for   An²+Bn+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  n =   ————————
                      2A

  In our case,  A   =     1
                      B   =    2
                      C   =  -1095

Accordingly,  B²  -  4AC   =
                     4 - (-4380) =
                     4384

Applying the quadratic formula :

               -2 ± √ 4384
   n  =    ——————
                      2

Can  √ 4384 be simplified ?

Yes!   The prime factorization of  4384   is
   2•2•2•2•2•137 
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).

√ 4384   =  √ 2•2•2•2•2•137   =2•2•√ 274   =
                ±  4 • √ 274

  √ 274   , rounded to 4 decimal digits, is  16.5529
 So now we are looking at:
           n  =  ( -2 ± 4 •  16.553 ) / 2

Two real solutions:

 n =(-2+√4384)/2=-1+2√ 274 = 32.106

or:

 n =(-2-√4384)/2=-1-2√ 274 = -34.106

Two solutions were found :

1.  n =(-2-√4384)/2=-1-2√ 274 = -34.106
2.  n =(-2+√4384)/2=-1+2√ 274 = 32.106