Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (24t2 -  90t) +  120  = 0 

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   16t² - 90t + 120  =   2 • (8t² - 45t + 60) 

Trying to factor by splitting the middle term

 3.2     Factoring  8t² - 45t + 60 

The first term is,  8t²  its coefficient is  8 .
The middle term is,  -45t  its coefficient is  -45 .
The last term, "the constant", is  +60 

Step-1 : Multiply the coefficient of the first term by the constant   8 • 60 = 480 

Step-2 : Find two factors of  480  whose sum equals the coefficient of the middle term, which is   -45 .

For tidiness, printing of 42 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 - 45t + 60)  = 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 = 8t²-45t+60

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   2.8125  

 Plugging into the parabola formula   2.8125  for  t  we can calculate the  y -coordinate : 
  y = 8.0 * 2.81 * 2.81 - 45.0 * 2.81 + 60.0
or   y = -3.281

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 8t²-45t+60
Axis of Symmetry (dashed)  {t}={ 2.81} 
Vertex at  {t,y} = { 2.81,-3.28} 
 t -Intercepts (Roots) :
Root 1 at  {t,y} = { 2.17, 0.00} 
Root 2 at  {t,y} = { 3.45, 0.00} 

Solve Quadratic Equation by Completing The Square

 4.3     Solving   8t²-45t+60 = 0 by Completing The Square .

 Divide both sides of the equation by  8  to have 1 as the coefficient of the first term :
   t²-(45/8)t+(15/2) = 0

Subtract  15/2  from both side of the equation :
   t²-(45/8)t = -15/2

Now the clever bit: Take the coefficient of  t , which is  45/8 , divide by two, giving  45/16 , and finally square it giving  2025/256 

Add  2025/256  to both sides of the equation :
  On the right hand side we have :
   -15/2  +  2025/256   The common denominator of the two fractions is  256   Adding  (-1920/256)+(2025/256)  gives  105/256 
  So adding to both sides we finally get :
   t²-(45/8)t+(2025/256) = 105/256

Adding  2025/256  has completed the left hand side into a perfect square :
   t²-(45/8)t+(2025/256)  =
   (t-(45/16)) • (t-(45/16))  =
  (t-(45/16))²
Things which are equal to the same thing are also equal to one another. Since
   t²-(45/8)t+(2025/256) = 105/256 and
   t²-(45/8)t+(2025/256) = (t-(45/16))²
then, according to the law of transitivity,
   (t-(45/16))² = 105/256

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-(45/16))²   is
   (t-(45/16))^2/2 =
  (t-(45/16))¹ =
   t-(45/16)

Now, applying the Square Root Principle to  Eq. #4.3.1  we get:
   t-(45/16) = √ 105/256

Add  45/16  to both sides to obtain:
   t = 45/16 + √ 105/256

Since a square root has two values, one positive and the other negative
   t² - (45/8)t + (15/2) = 0
   has two solutions:
  t = 45/16 + √ 105/256
   or
  t = 45/16 - √ 105/256

Note that  √ 105/256 can be written as
  √ 105  / √ 256   which is √ 105  / 16

Solve Quadratic Equation using the Quadratic Formula

 4.4     Solving    8t²-45t+60 = 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   =   -45
                      C   =   60

Accordingly,  B²  -  4AC   =
                     2025 - 1920 =
                     105

Applying the quadratic formula :

               45 ± √ 105
   t  =    ——————
                      16

  √ 105   , rounded to 4 decimal digits, is  10.2470
 So now we are looking at:
           t  =  ( 45 ±  10.247 ) / 16

Two real solutions:

 t =(45+√105)/16= 3.453

or:

 t =(45-√105)/16= 2.172

Two solutions were found :

1.  t =(45-√105)/16= 2.172
2.  t =(45+√105)/16= 3.453