Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((0 -  (28•32x2)) +  288x) +  1  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  -2304x²+288x+1 

The first term is,  -2304x²  its coefficient is  -2304 .
The middle term is,  +288x  its coefficient is  288 .
The last term, "the constant", is  +1 

Step-1 : Multiply the coefficient of the first term by the constant   -2304 • 1 = -2304 

Step-2 : Find two factors of  -2304  whose sum equals the coefficient of the middle term, which is   288 .

For tidiness, printing of 21 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  2  :

  -2304x2 + 288x + 1  = 0 

Step  3  :

Parabola, Finding the Vertex :

 3.1      Find the Vertex of   y = -2304x²+288x+1

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens down and accordingly has a highest point (AKA absolute maximum) .    We know this even before plotting  "y"  because the coefficient of the first term, -2304 , is negative (smaller 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   0.0625  

 Plugging into the parabola formula   0.0625  for  x  we can calculate the  y -coordinate : 
  y = -2304.0 * 0.06 * 0.06 + 288.0 * 0.06 + 1.0
or   y = 10.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = -2304x²+288x+1
Axis of Symmetry (dashed)  {x}={ 0.06} 
Vertex at  {x,y} = { 0.06,10.00} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = { 0.13, 0.00} 
Root 2 at  {x,y} = {-0.00, 0.00} 

Solve Quadratic Equation by Completing The Square

 3.2     Solving   -2304x²+288x+1 = 0 by Completing The Square .

 Multiply both sides of the equation by  (-1)  to obtain positive coefficient for the first term:
 2304x²-288x-1 = 0  Divide both sides of the equation by  2304  to have 1 as the coefficient of the first term :
   x²-(1/8)x-(1/2304) = 0

Add  1/2304  to both side of the equation :
   x²-(1/8)x = 1/2304

Now the clever bit: Take the coefficient of  x , which is  1/8 , divide by two, giving  1/16 , and finally square it giving  1/256 

Add  1/256  to both sides of the equation :
  On the right hand side we have :
   1/2304  +  1/256   The common denominator of the two fractions is  2304   Adding  (1/2304)+(9/2304)  gives  10/2304 
  So adding to both sides we finally get :
   x²-(1/8)x+(1/256) = 5/1152

Adding  1/256  has completed the left hand side into a perfect square :
   x²-(1/8)x+(1/256)  =
   (x-(1/16)) • (x-(1/16))  =
  (x-(1/16))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(1/8)x+(1/256) = 5/1152 and
   x²-(1/8)x+(1/256) = (x-(1/16))²
then, according to the law of transitivity,
   (x-(1/16))² = 5/1152

We'll refer to this Equation as  Eq. #3.2.1  

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

Note that the square root of
   (x-(1/16))²   is
   (x-(1/16))^2/2 =
  (x-(1/16))¹ =
   x-(1/16)

Now, applying the Square Root Principle to  Eq. #3.2.1  we get:
   x-(1/16) = √ 5/1152

Add  1/16  to both sides to obtain:
   x = 1/16 + √ 5/1152

Since a square root has two values, one positive and the other negative
   x² - (1/8)x - (1/2304) = 0
   has two solutions:
  x = 1/16 + √ 5/1152
   or
  x = 1/16 - √ 5/1152

Note that  √ 5/1152 can be written as
  √ 5  / √ 1152 

Solve Quadratic Equation using the Quadratic Formula

 3.3     Solving    -2304x²+288x+1 = 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   =    -2304
                      B   =   288
                      C   =   1

Accordingly,  B²  -  4AC   =
                     82944 - (-9216) =
                     92160

Applying the quadratic formula :

               -288 ± √ 92160
   x  =    ————————
                        -4608

Can  √ 92160 be simplified ?

Yes!   The prime factorization of  92160   is
   2•2•2•2•2•2•2•2•2•2•2•3•3•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).

√ 92160   =  √ 2•2•2•2•2•2•2•2•2•2•2•3•3•5   =2•2•2•2•2•3•√ 10   =
                ±  96 • √ 10

  √ 10   , rounded to 4 decimal digits, is   3.1623
 So now we are looking at:
           x  =  ( -288 ± 96 •  3.162 ) / -4608

Two real solutions:

 x =(-288+√92160)/-4608=1/16-1/48√ 10 = -0.003

or:

 x =(-288-√92160)/-4608=1/16+1/48√ 10 = 0.128

Two solutions were found :

1.  x =(-288-√92160)/-4608=1/16+1/48√ 10 = 0.128
2.  x =(-288+√92160)/-4608=1/16-1/48√ 10 = -0.003