Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (3t2 -  90t) +  500  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  3t²-90t+500 

The first term is,  3t²  its coefficient is  3 .
The middle term is,  -90t  its coefficient is  -90 .
The last term, "the constant", is  +500 

Step-1 : Multiply the coefficient of the first term by the constant   3 • 500 = 1500 

Step-2 : Find two factors of  1500  whose sum equals the coefficient of the middle term, which is   -90 .

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  2  :

  3t2 - 90t + 500  = 0 

Step  3  :

Parabola, Finding the Vertex :

 3.1      Find the Vertex of   y = 3t²-90t+500

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  15.0000  

 Plugging into the parabola formula  15.0000  for  t  we can calculate the  y -coordinate : 
  y = 3.0 * 15.00 * 15.00 - 90.0 * 15.00 + 500.0
or   y = -175.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 3t²-90t+500
Axis of Symmetry (dashed)  {t}={15.00} 
Vertex at  {t,y} = {15.00,-175.00} 
 t -Intercepts (Roots) :
Root 1 at  {t,y} = { 7.36, 0.00} 
Root 2 at  {t,y} = {22.64, 0.00} 

Solve Quadratic Equation by Completing The Square

 3.2     Solving   3t²-90t+500 = 0 by Completing The Square .

 Divide both sides of the equation by  3  to have 1 as the coefficient of the first term :
   t²-30t+(500/3) = 0

Subtract  500/3  from both side of the equation :
   t²-30t = -500/3

Now the clever bit: Take the coefficient of  t , which is  30 , divide by two, giving  15 , and finally square it giving  225 

Add  225  to both sides of the equation :
  On the right hand side we have :
   -500/3  +  225    or,  (-500/3)+(225/1) 
  The common denominator of the two fractions is  3   Adding  (-500/3)+(675/3)  gives  175/3 
  So adding to both sides we finally get :
   t²-30t+225 = 175/3

Adding  225  has completed the left hand side into a perfect square :
   t²-30t+225  =
   (t-15) • (t-15)  =
  (t-15)²
Things which are equal to the same thing are also equal to one another. Since
   t²-30t+225 = 175/3 and
   t²-30t+225 = (t-15)²
then, according to the law of transitivity,
   (t-15)² = 175/3

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-15)²   is
   (t-15)^2/2 =
  (t-15)¹ =
   t-15

Now, applying the Square Root Principle to  Eq. #3.2.1  we get:
   t-15 = √ 175/3

Add  15  to both sides to obtain:
   t = 15 + √ 175/3

Since a square root has two values, one positive and the other negative
   t² - 30t + (500/3) = 0
   has two solutions:
  t = 15 + √ 175/3
   or
  t = 15 - √ 175/3

Note that  √ 175/3 can be written as
  √ 175  / √ 3 

Solve Quadratic Equation using the Quadratic Formula

 3.3     Solving    3t²-90t+500 = 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   =   -90
                      C   =  500

Accordingly,  B²  -  4AC   =
                     8100 - 6000 =
                     2100

Applying the quadratic formula :

               90 ± √ 2100
   t  =    ——————
                      6

Can  √ 2100 be simplified ?

Yes!   The prime factorization of  2100   is
   2•2•3•5•5•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).

√ 2100   =  √ 2•2•3•5•5•7   =2•5•√ 21   =
                ±  10 • √ 21

  √ 21   , rounded to 4 decimal digits, is   4.5826
 So now we are looking at:
           t  =  ( 90 ± 10 •  4.583 ) / 6

Two real solutions:

 t =(90+√2100)/6=15+5/3√ 21 = 22.638

or:

 t =(90-√2100)/6=15-5/3√ 21 = 7.362

Two solutions were found :

1.  t =(90-√2100)/6=15-5/3√ 21 = 7.362
2.  t =(90+√2100)/6=15+5/3√ 21 = 22.638