Solution - Quadratic equations
Step by Step Solution
Step by step solution :
Step 1 :
Trying to factor by splitting the middle term
 1.1     Factoring  x2-28x+36 
 The first term is,  x2  its coefficient is  1 .
The middle term is,  -28x  its coefficient is  -28 .
The last term, "the constant", is  +36 
Step-1 : Multiply the coefficient of the first term by the constant   1 • 36 = 36 
Step-2 : Find two factors of  36  whose sum equals the coefficient of the middle term, which is   -28 .
| -36 | + | -1 | = | -37 | ||
| -18 | + | -2 | = | -20 | ||
| -12 | + | -3 | = | -15 | ||
| -9 | + | -4 | = | -13 | ||
| -6 | + | -6 | = | -12 | ||
| -4 | + | -9 | = | -13 | 
For tidiness, printing of 12 lines which failed to find two such factors, was suppressed
Observation : No two such factors can be found !! 
 Conclusion : Trinomial can not be factored 
Equation at the end of step 1 :
  x2 - 28x + 36  = 0 
Step 2 :
Parabola, Finding the Vertex :
 2.1      Find the Vertex of   y = x2-28x+36
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,Ax2+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is  14.0000  
 Plugging into the parabola formula  14.0000  for  x  we can calculate the  y -coordinate : 
  y = 1.0 * 14.00 * 14.00 - 28.0 * 14.00 + 36.0 
 or    y = -160.000
Parabola, Graphing Vertex and X-Intercepts :
Root plot for :  y = x2-28x+36
 Axis of Symmetry (dashed)  {x}={14.00} 
 Vertex at  {x,y} = {14.00,-160.00}  
  x -Intercepts (Roots) :
 Root 1 at  {x,y} = { 1.35, 0.00} 
 Root 2 at  {x,y} = {26.65, 0.00} 
 
Solve Quadratic Equation by Completing The Square
 2.2     Solving   x2-28x+36 = 0 by Completing The Square .
 Subtract  36  from both side of the equation :
   x2-28x = -36
Now the clever bit: Take the coefficient of  x , 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 :
   -36  +  196    or,  (-36/1)+(196/1) 
  The common denominator of the two fractions is  1   Adding  (-36/1)+(196/1)  gives  160/1 
  So adding to both sides we finally get :
   x2-28x+196 = 160
Adding  196  has completed the left hand side into a perfect square :
   x2-28x+196  =
   (x-14) • (x-14)  =
  (x-14)2 
Things which are equal to the same thing are also equal to one another. Since
   x2-28x+196 = 160 and
   x2-28x+196 = (x-14)2 
then, according to the law of transitivity,
   (x-14)2 = 160
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
   (x-14)2   is
   (x-14)2/2 =
  (x-14)1 =
   x-14
Now, applying the Square Root Principle to  Eq. #2.2.1  we get:
   x-14 = √ 160 
Add  14  to both sides to obtain:
   x = 14 + √ 160 
Since a square root has two values, one positive and the other negative
   x2 - 28x + 36 = 0
   has two solutions:
  x = 14 + √ 160 
   or
  x = 14 - √ 160 
Solve Quadratic Equation using the Quadratic Formula
 2.3     Solving    x2-28x+36 = 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   =     1
                      B   =   -28
                      C   =   36 
Accordingly,  B2  -  4AC   =
                     784 - 144 =
                     640
Applying the quadratic formula :
                28 ± √ 640 
   x  =    ——————
                      2
Can  √ 640  be simplified ?
Yes!   The prime factorization of  640   is
   2•2•2•2•2•2•2•5  
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).
√ 640   =  √ 2•2•2•2•2•2•2•5   =2•2•2•√ 10   =
                ±  8 • √ 10 
  √ 10   , rounded to 4 decimal digits, is   3.1623
 So now we are looking at:
           x  =  ( 28 ± 8 •  3.162 ) / 2
Two real solutions:
 x =(28+√640)/2=14+4√ 10 = 26.649 
or:
 x =(28-√640)/2=14-4√ 10 = 1.351 
Two solutions were found :
- x =(28-√640)/2=14-4√ 10 = 1.351
 - x =(28+√640)/2=14+4√ 10 = 26.649
 
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