Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (2t2 +  40t) -  300  = 0 

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   2t² + 40t - 300  =   2 • (t² + 20t - 150) 

Trying to factor by splitting the middle term

 3.2     Factoring  t² + 20t - 150 

The first term is,  t²  its coefficient is  1 .
The middle term is,  +20t  its coefficient is  20 .
The last term, "the constant", is  -150 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -150 = -150 

Step-2 : Find two factors of  -150  whose sum equals the coefficient of the middle term, which is   20 .

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

Equation at the end of step  3  :

  2 • (t2 + 20t - 150)  = 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 = t²+20t-150

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  -10.0000  

 Plugging into the parabola formula  -10.0000  for  t  we can calculate the  y -coordinate : 
  y = 1.0 * -10.00 * -10.00 + 20.0 * -10.00 - 150.0
or   y = -250.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = t²+20t-150
Axis of Symmetry (dashed)  {t}={-10.00} 
Vertex at  {t,y} = {-10.00,-250.00} 
 t -Intercepts (Roots) :
Root 1 at  {t,y} = {-25.81, 0.00} 
Root 2 at  {t,y} = { 5.81, 0.00} 

Solve Quadratic Equation by Completing The Square

 4.3     Solving   t²+20t-150 = 0 by Completing The Square .

 Add  150  to both side of the equation :
   t²+20t = 150

Now the clever bit: Take the coefficient of  t , which is  20 , divide by two, giving  10 , and finally square it giving  100 

Add  100  to both sides of the equation :
  On the right hand side we have :
   150  +  100    or,  (150/1)+(100/1) 
  The common denominator of the two fractions is  1   Adding  (150/1)+(100/1)  gives  250/1 
  So adding to both sides we finally get :
   t²+20t+100 = 250

Adding  100  has completed the left hand side into a perfect square :
   t²+20t+100  =
   (t+10) • (t+10)  =
  (t+10)²
Things which are equal to the same thing are also equal to one another. Since
   t²+20t+100 = 250 and
   t²+20t+100 = (t+10)²
then, according to the law of transitivity,
   (t+10)² = 250

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+10)²   is
   (t+10)^2/2 =
  (t+10)¹ =
   t+10

Now, applying the Square Root Principle to  Eq. #4.3.1  we get:
   t+10 = √ 250

Subtract  10  from both sides to obtain:
   t = -10 + √ 250

Since a square root has two values, one positive and the other negative
   t² + 20t - 150 = 0
   has two solutions:
  t = -10 + √ 250
   or
  t = -10 - √ 250

Solve Quadratic Equation using the Quadratic Formula

 4.4     Solving    t²+20t-150 = 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   =    20
                      C   =  -150

Accordingly,  B²  -  4AC   =
                     400 - (-600) =
                     1000

Applying the quadratic formula :

               -20 ± √ 1000
   t  =    ——————
                      2

Can  √ 1000 be simplified ?

Yes!   The prime factorization of  1000   is
   2•2•2•5•5•5 
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).

√ 1000   =  √ 2•2•2•5•5•5   =2•5•√ 10   =
                ±  10 • √ 10

  √ 10   , rounded to 4 decimal digits, is   3.1623
 So now we are looking at:
           t  =  ( -20 ± 10 •  3.162 ) / 2

Two real solutions:

 t =(-20+√1000)/2=-10+5√ 10 = 5.811

or:

 t =(-20-√1000)/2=-10-5√ 10 = -25.811

Two solutions were found :

1.  t =(-20-√1000)/2=-10-5√ 10 = -25.811
2.  t =(-20+√1000)/2=-10+5√ 10 = 5.811