Solution - Quadratic equations

Step by step solution :

Step  1  :

Trying to factor by splitting the middle term

 1.1     Factoring  t²+50t-200 

The first term is,  t²  its coefficient is  1 .
The middle term is,  +50t  its coefficient is  50 .
The last term, "the constant", is  -200 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -200 = -200 

Step-2 : Find two factors of  -200  whose sum equals the coefficient of the middle term, which is   50 .

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

Equation at the end of step  1  :

  t2 + 50t - 200  = 0 

Step  2  :

Parabola, Finding the Vertex :

 2.1      Find the Vertex of   y = t²+50t-200

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, 1 , 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  -25.0000  

 Plugging into the parabola formula  -25.0000  for  t  we can calculate the  y -coordinate : 
  y = 1.0 * -25.00 * -25.00 + 50.0 * -25.00 - 200.0
or   y = -825.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = t²+50t-200
Axis of Symmetry (dashed)  {t}={-25.00} 
Vertex at  {t,y} = {-25.00,-825.00} 
 t -Intercepts (Roots) :
Root 1 at  {t,y} = {-53.72, 0.00} 
Root 2 at  {t,y} = { 3.72, 0.00} 

Solve Quadratic Equation by Completing The Square

 2.2     Solving   t²+50t-200 = 0 by Completing The Square .

 Add  200  to both side of the equation :
   t²+50t = 200

Now the clever bit: Take the coefficient of  t , which is  50 , divide by two, giving  25 , and finally square it giving  625 

Add  625  to both sides of the equation :
  On the right hand side we have :
   200  +  625    or,  (200/1)+(625/1) 
  The common denominator of the two fractions is  1   Adding  (200/1)+(625/1)  gives  825/1 
  So adding to both sides we finally get :
   t²+50t+625 = 825

Adding  625  has completed the left hand side into a perfect square :
   t²+50t+625  =
   (t+25) • (t+25)  =
  (t+25)²
Things which are equal to the same thing are also equal to one another. Since
   t²+50t+625 = 825 and
   t²+50t+625 = (t+25)²
then, according to the law of transitivity,
   (t+25)² = 825

We'll refer to this Equation as  Eq. #2.2.1  

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

Note that the square root of
   (t+25)²   is
   (t+25)^2/2 =
  (t+25)¹ =
   t+25

Now, applying the Square Root Principle to  Eq. #2.2.1  we get:
   t+25 = √ 825

Subtract  25  from both sides to obtain:
   t = -25 + √ 825

Since a square root has two values, one positive and the other negative
   t² + 50t - 200 = 0
   has two solutions:
  t = -25 + √ 825
   or
  t = -25 - √ 825

Solve Quadratic Equation using the Quadratic Formula

 2.3     Solving    t²+50t-200 = 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   =     1
                      B   =    50
                      C   =  -200

Accordingly,  B²  -  4AC   =
                     2500 - (-800) =
                     3300

Applying the quadratic formula :

               -50 ± √ 3300
   t  =    ———————
                        2

Can  √ 3300 be simplified ?

Yes!   The prime factorization of  3300   is
   2•2•3•5•5•11 
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).

√ 3300   =  √ 2•2•3•5•5•11   =2•5•√ 33   =
                ±  10 • √ 33

  √ 33   , rounded to 4 decimal digits, is   5.7446
 So now we are looking at:
           t  =  ( -50 ± 10 •  5.745 ) / 2

Two real solutions:

 t =(-50+√3300)/2=-25+5√ 33 = 3.723

or:

 t =(-50-√3300)/2=-25-5√ 33 = -53.723

Two solutions were found :

1.  t =(-50-√3300)/2=-25-5√ 33 = -53.723
2.  t =(-50+√3300)/2=-25+5√ 33 = 3.723