Solution - Quadratic equations

Step by step solution :

Step  1  :

Trying to factor by splitting the middle term

 1.1     Factoring  w²-24w-13 

The first term is,  w²  its coefficient is  1 .
The middle term is,  -24w  its coefficient is  -24 .
The last term, "the constant", is  -13 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -13 = -13 

Step-2 : Find two factors of  -13  whose sum equals the coefficient of the middle term, which is   -24 .

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

Equation at the end of step  1  :

  w2 - 24w - 13  = 0 

Step  2  :

Parabola, Finding the Vertex :

 2.1      Find the Vertex of   y = w²-24w-13

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,Aw²+Bw+C,the  w -coordinate of the vertex is given by  -B/(2A) . In our case the  w  coordinate is  12.0000  

 Plugging into the parabola formula  12.0000  for  w  we can calculate the  y -coordinate : 
  y = 1.0 * 12.00 * 12.00 - 24.0 * 12.00 - 13.0
or   y = -157.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = w²-24w-13
Axis of Symmetry (dashed)  {w}={12.00} 
Vertex at  {w,y} = {12.00,-157.00} 
 w -Intercepts (Roots) :
Root 1 at  {w,y} = {-0.53, 0.00} 
Root 2 at  {w,y} = {24.53, 0.00} 

Solve Quadratic Equation by Completing The Square

 2.2     Solving   w²-24w-13 = 0 by Completing The Square .

 Add  13  to both side of the equation :
   w²-24w = 13

Now the clever bit: Take the coefficient of  w , which is  24 , divide by two, giving  12 , and finally square it giving  144 

Add  144  to both sides of the equation :
  On the right hand side we have :
   13  +  144    or,  (13/1)+(144/1) 
  The common denominator of the two fractions is  1   Adding  (13/1)+(144/1)  gives  157/1 
  So adding to both sides we finally get :
   w²-24w+144 = 157

Adding  144  has completed the left hand side into a perfect square :
   w²-24w+144  =
   (w-12) • (w-12)  =
  (w-12)²
Things which are equal to the same thing are also equal to one another. Since
   w²-24w+144 = 157 and
   w²-24w+144 = (w-12)²
then, according to the law of transitivity,
   (w-12)² = 157

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
   (w-12)²   is
   (w-12)^2/2 =
  (w-12)¹ =
   w-12

Now, applying the Square Root Principle to  Eq. #2.2.1  we get:
   w-12 = √ 157

Add  12  to both sides to obtain:
   w = 12 + √ 157

Since a square root has two values, one positive and the other negative
   w² - 24w - 13 = 0
   has two solutions:
  w = 12 + √ 157
   or
  w = 12 - √ 157

Solve Quadratic Equation using the Quadratic Formula

 2.3     Solving    w²-24w-13 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  w  , the solution for   Aw²+Bw+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  w =   ————————
                      2A

  In our case,  A   =     1
                      B   =   -24
                      C   =  -13

Accordingly,  B²  -  4AC   =
                     576 - (-52) =
                     628

Applying the quadratic formula :

               24 ± √ 628
   w  =    ——————
                      2

Can  √ 628 be simplified ?

Yes!   The prime factorization of  628   is
   2•2•157 
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).

√ 628   =  √ 2•2•157   =
                ±  2 • √ 157

  √ 157   , rounded to 4 decimal digits, is  12.5300
 So now we are looking at:
           w  =  ( 24 ± 2 •  12.530 ) / 2

Two real solutions:

 w =(24+√628)/2=12+√ 157 = 24.530

or:

 w =(24-√628)/2=12-√ 157 = -0.530

Two solutions were found :

1.  w =(24-√628)/2=12-√ 157 = -0.530
2.  w =(24+√628)/2=12+√ 157 = 24.530