Solution - Quadratic equations

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                     0-(157-9*t-16*t^2)=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  0 -  ((157 - 9t) -  24t2)  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  16t²+9t-157 

The first term is,  16t²  its coefficient is  16 .
The middle term is,  +9t  its coefficient is  9 .
The last term, "the constant", is  -157 

Step-1 : Multiply the coefficient of the first term by the constant   16 • -157 = -2512 

Step-2 : Find two factors of  -2512  whose sum equals the coefficient of the middle term, which is   9 .

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

Equation at the end of step  2  :

  16t2 + 9t - 157  = 0 

Step  3  :

Parabola, Finding the Vertex :

 3.1      Find the Vertex of   y = 16t²+9t-157

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, 16 , 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  -0.2812  

 Plugging into the parabola formula  -0.2812  for  t  we can calculate the  y -coordinate : 
  y = 16.0 * -0.28 * -0.28 + 9.0 * -0.28 - 157.0
or   y = -158.266

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 16t²+9t-157
Axis of Symmetry (dashed)  {t}={-0.28} 
Vertex at  {t,y} = {-0.28,-158.27} 
 t -Intercepts (Roots) :
Root 1 at  {t,y} = {-3.43, 0.00} 
Root 2 at  {t,y} = { 2.86, 0.00} 

Solve Quadratic Equation by Completing The Square

 3.2     Solving   16t²+9t-157 = 0 by Completing The Square .

 Divide both sides of the equation by  16  to have 1 as the coefficient of the first term :
   t²+(9/16)t-(157/16) = 0

Add  157/16  to both side of the equation :
   t²+(9/16)t = 157/16

Now the clever bit: Take the coefficient of  t , which is  9/16 , divide by two, giving  9/32 , and finally square it giving  81/1024 

Add  81/1024  to both sides of the equation :
  On the right hand side we have :
   157/16  +  81/1024   The common denominator of the two fractions is  1024   Adding  (10048/1024)+(81/1024)  gives  10129/1024 
  So adding to both sides we finally get :
   t²+(9/16)t+(81/1024) = 10129/1024

Adding  81/1024  has completed the left hand side into a perfect square :
   t²+(9/16)t+(81/1024)  =
   (t+(9/32)) • (t+(9/32))  =
  (t+(9/32))²
Things which are equal to the same thing are also equal to one another. Since
   t²+(9/16)t+(81/1024) = 10129/1024 and
   t²+(9/16)t+(81/1024) = (t+(9/32))²
then, according to the law of transitivity,
   (t+(9/32))² = 10129/1024

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+(9/32))²   is
   (t+(9/32))^2/2 =
  (t+(9/32))¹ =
   t+(9/32)

Now, applying the Square Root Principle to  Eq. #3.2.1  we get:
   t+(9/32) = √ 10129/1024

Subtract  9/32  from both sides to obtain:
   t = -9/32 + √ 10129/1024

Since a square root has two values, one positive and the other negative
   t² + (9/16)t - (157/16) = 0
   has two solutions:
  t = -9/32 + √ 10129/1024
   or
  t = -9/32 - √ 10129/1024

Note that  √ 10129/1024 can be written as
  √ 10129  / √ 1024   which is √ 10129  / 32

Solve Quadratic Equation using the Quadratic Formula

 3.3     Solving    16t²+9t-157 = 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   =     16
                      B   =    9
                      C   =  -157

Accordingly,  B²  -  4AC   =
                     81 - (-10048) =
                     10129

Applying the quadratic formula :

               -9 ± √ 10129
   t  =    ———————
                        32

  √ 10129   , rounded to 4 decimal digits, is  100.6429
 So now we are looking at:
           t  =  ( -9 ±  100.643 ) / 32

Two real solutions:

 t =(-9+√10129)/32= 2.864

or:

 t =(-9-√10129)/32=-3.426

Two solutions were found :

1.  t =(-9-√10129)/32=-3.426
2.  t =(-9+√10129)/32= 2.864