Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((0 -  24t2) -  2t) +  500  = 0 

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   -16t² - 2t + 500  =   -2 • (8t² + t - 250) 

Trying to factor by splitting the middle term

 3.2     Factoring  8t² + t - 250 

The first term is,  8t²  its coefficient is  8 .
The middle term is,  +t  its coefficient is  1 .
The last term, "the constant", is  -250 

Step-1 : Multiply the coefficient of the first term by the constant   8 • -250 = -2000 

Step-2 : Find two factors of  -2000  whose sum equals the coefficient of the middle term, which is   1 .

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  3  :

  -2 • (8t2 + t - 250)  = 0 

Step  4  :

Equations which are never true :

 4.1      Solve :    -2   =  0

This equation has no solution.
A a non-zero constant never equals zero.

Parabola, Finding the Vertex :

 4.2      Find the Vertex of   y = 8t²+t-250

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, 8 , 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  -0.0625  

 Plugging into the parabola formula  -0.0625  for  t  we can calculate the  y -coordinate : 
  y = 8.0 * -0.06 * -0.06 + 1.0 * -0.06 - 250.0
or   y = -250.031

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 8t²+t-250
Axis of Symmetry (dashed)  {t}={-0.06} 
Vertex at  {t,y} = {-0.06,-250.03} 
 t -Intercepts (Roots) :
Root 1 at  {t,y} = {-5.65, 0.00} 
Root 2 at  {t,y} = { 5.53, 0.00} 

Solve Quadratic Equation by Completing The Square

 4.3     Solving   8t²+t-250 = 0 by Completing The Square .

 Divide both sides of the equation by  8  to have 1 as the coefficient of the first term :
   t²+(1/8)t-(125/4) = 0

Add  125/4  to both side of the equation :
   t²+(1/8)t = 125/4

Now the clever bit: Take the coefficient of  t , 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 :
   125/4  +  1/256   The common denominator of the two fractions is  256   Adding  (8000/256)+(1/256)  gives  8001/256 
  So adding to both sides we finally get :
   t²+(1/8)t+(1/256) = 8001/256

Adding  1/256  has completed the left hand side into a perfect square :
   t²+(1/8)t+(1/256)  =
   (t+(1/16)) • (t+(1/16))  =
  (t+(1/16))²
Things which are equal to the same thing are also equal to one another. Since
   t²+(1/8)t+(1/256) = 8001/256 and
   t²+(1/8)t+(1/256) = (t+(1/16))²
then, according to the law of transitivity,
   (t+(1/16))² = 8001/256

We'll refer to this Equation as  Eq. #4.3.1  

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

Note that the square root of
   (t+(1/16))²   is
   (t+(1/16))^2/2 =
  (t+(1/16))¹ =
   t+(1/16)

Now, applying the Square Root Principle to  Eq. #4.3.1  we get:
   t+(1/16) = √ 8001/256

Subtract  1/16  from both sides to obtain:
   t = -1/16 + √ 8001/256

Since a square root has two values, one positive and the other negative
   t² + (1/8)t - (125/4) = 0
   has two solutions:
  t = -1/16 + √ 8001/256
   or
  t = -1/16 - √ 8001/256

Note that  √ 8001/256 can be written as
  √ 8001  / √ 256   which is √ 8001  / 16

Solve Quadratic Equation using the Quadratic Formula

 4.4     Solving    8t²+t-250 = 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   =     8
                      B   =    1
                      C   =  -250

Accordingly,  B²  -  4AC   =
                     1 - (-8000) =
                     8001

Applying the quadratic formula :

               -1 ± √ 8001
   t  =    ——————
                      16

Can  √ 8001 be simplified ?

Yes!   The prime factorization of  8001   is
   3•3•7•127 
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).

√ 8001   =  √ 3•3•7•127   =
                ±  3 • √ 889

  √ 889   , rounded to 4 decimal digits, is  29.8161
 So now we are looking at:
           t  =  ( -1 ± 3 •  29.816 ) / 16

Two real solutions:

 t =(-1+√8001)/16=(-1+3√ 889 )/16= 5.528

or:

 t =(-1-√8001)/16=(-1-3√ 889 )/16= -5.653

Two solutions were found :

1.  t =(-1-√8001)/16=(-1-3√ 889 )/16= -5.653
2.  t =(-1+√8001)/16=(-1+3√ 889 )/16= 5.528