Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (3t2 -  32t) -  512  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  3t²-32t-512 

The first term is,  3t²  its coefficient is  3 .
The middle term is,  -32t  its coefficient is  -32 .
The last term, "the constant", is  -512 

Step-1 : Multiply the coefficient of the first term by the constant   3 • -512 = -1536 

Step-2 : Find two factors of  -1536  whose sum equals the coefficient of the middle term, which is   -32 .

For tidiness, printing of 14 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  :

  3t2 - 32t - 512  = 0 

Step  3  :

Parabola, Finding the Vertex :

 3.1      Find the Vertex of   y = 3t²-32t-512

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, 3 , 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,At²+Bt+C,the  t -coordinate of the vertex is given by  -B/(2A) . In our case the  t  coordinate is   5.3333  

 Plugging into the parabola formula   5.3333  for  t  we can calculate the  y -coordinate : 
  y = 3.0 * 5.33 * 5.33 - 32.0 * 5.33 - 512.0
or   y = -597.333

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 3t²-32t-512
Axis of Symmetry (dashed)  {t}={ 5.33} 
Vertex at  {t,y} = { 5.33,-597.33} 
 t -Intercepts (Roots) :
Root 1 at  {t,y} = {-8.78, 0.00} 
Root 2 at  {t,y} = {19.44, 0.00} 

Solve Quadratic Equation by Completing The Square

 3.2     Solving   3t²-32t-512 = 0 by Completing The Square .

 Divide both sides of the equation by  3  to have 1 as the coefficient of the first term :
   t²-(32/3)t-(512/3) = 0

Add  512/3  to both side of the equation :
   t²-(32/3)t = 512/3

Now the clever bit: Take the coefficient of  t , which is  32/3 , divide by two, giving  16/3 , and finally square it giving  256/9 

Add  256/9  to both sides of the equation :
  On the right hand side we have :
   512/3  +  256/9   The common denominator of the two fractions is  9   Adding  (1536/9)+(256/9)  gives  1792/9 
  So adding to both sides we finally get :
   t²-(32/3)t+(256/9) = 1792/9

Adding  256/9  has completed the left hand side into a perfect square :
   t²-(32/3)t+(256/9)  =
   (t-(16/3)) • (t-(16/3))  =
  (t-(16/3))²
Things which are equal to the same thing are also equal to one another. Since
   t²-(32/3)t+(256/9) = 1792/9 and
   t²-(32/3)t+(256/9) = (t-(16/3))²
then, according to the law of transitivity,
   (t-(16/3))² = 1792/9

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
   (t-(16/3))²   is
   (t-(16/3))^2/2 =
  (t-(16/3))¹ =
   t-(16/3)

Now, applying the Square Root Principle to  Eq. #3.2.1  we get:
   t-(16/3) = √ 1792/9

Add  16/3  to both sides to obtain:
   t = 16/3 + √ 1792/9

Since a square root has two values, one positive and the other negative
   t² - (32/3)t - (512/3) = 0
   has two solutions:
  t = 16/3 + √ 1792/9
   or
  t = 16/3 - √ 1792/9

Note that  √ 1792/9 can be written as
  √ 1792  / √ 9   which is √ 1792  / 3

Solve Quadratic Equation using the Quadratic Formula

 3.3     Solving    3t²-32t-512 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  t  , the solution for   At²+Bt+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  t =   ————————
                      2A

  In our case,  A   =     3
                      B   =   -32
                      C   =  -512

Accordingly,  B²  -  4AC   =
                     1024 - (-6144) =
                     7168

Applying the quadratic formula :

               32 ± √ 7168
   t  =    ——————
                      6

Can  √ 7168 be simplified ?

Yes!   The prime factorization of  7168   is
   2•2•2•2•2•2•2•2•2•2•7 
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).

√ 7168   =  √ 2•2•2•2•2•2•2•2•2•2•7   =2•2•2•2•2•√ 7   =
                ±  32 • √ 7

  √ 7   , rounded to 4 decimal digits, is   2.6458
 So now we are looking at:
           t  =  ( 32 ± 32 •  2.646 ) / 6

Two real solutions:

 t =(32+√7168)/6=(16+16√ 7 )/3= 19.444

or:

 t =(32-√7168)/6=(16-16√ 7 )/3= -8.777

Two solutions were found :

1.  t =(32-√7168)/6=(16-16√ 7 )/3= -8.777
2.  t =(32+√7168)/6=(16+16√ 7 )/3= 19.444