Solution - Quadratic equations

Rearrange:

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

                     100*x^2+60*x+9-(28)=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (((22•52x2) +  60x) +  9) -  28  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  100x²+60x-19 

The first term is,  100x²  its coefficient is  100 .
The middle term is,  +60x  its coefficient is  60 .
The last term, "the constant", is  -19 

Step-1 : Multiply the coefficient of the first term by the constant   100 • -19 = -1900 

Step-2 : Find two factors of  -1900  whose sum equals the coefficient of the middle term, which is   60 .

For tidiness, printing of 12 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  :

  100x2 + 60x - 19  = 0 

Step  3  :

Parabola, Finding the Vertex :

 3.1      Find the Vertex of   y = 100x²+60x-19

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, 100 , 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,Ax²+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is  -0.3000  

 Plugging into the parabola formula  -0.3000  for  x  we can calculate the  y -coordinate : 
  y = 100.0 * -0.30 * -0.30 + 60.0 * -0.30 - 19.0
or   y = -28.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 100x²+60x-19
Axis of Symmetry (dashed)  {x}={-0.30} 
Vertex at  {x,y} = {-0.30,-28.00} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-0.83, 0.00} 
Root 2 at  {x,y} = { 0.23, 0.00} 

Solve Quadratic Equation by Completing The Square

 3.2     Solving   100x²+60x-19 = 0 by Completing The Square .

 Divide both sides of the equation by  100  to have 1 as the coefficient of the first term :
   x²+(3/5)x-(19/100) = 0

Add  19/100  to both side of the equation :
   x²+(3/5)x = 19/100

Now the clever bit: Take the coefficient of  x , which is  3/5 , divide by two, giving  3/10 , and finally square it giving  9/100 

Add  9/100  to both sides of the equation :
  On the right hand side we have :
   19/100  +  9/100   The common denominator of the two fractions is  100   Adding  (19/100)+(9/100)  gives  28/100 
  So adding to both sides we finally get :
   x²+(3/5)x+(9/100) = 7/25

Adding  9/100  has completed the left hand side into a perfect square :
   x²+(3/5)x+(9/100)  =
   (x+(3/10)) • (x+(3/10))  =
  (x+(3/10))²
Things which are equal to the same thing are also equal to one another. Since
   x²+(3/5)x+(9/100) = 7/25 and
   x²+(3/5)x+(9/100) = (x+(3/10))²
then, according to the law of transitivity,
   (x+(3/10))² = 7/25

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
   (x+(3/10))²   is
   (x+(3/10))^2/2 =
  (x+(3/10))¹ =
   x+(3/10)

Now, applying the Square Root Principle to  Eq. #3.2.1  we get:
   x+(3/10) = √ 7/25

Subtract  3/10  from both sides to obtain:
   x = -3/10 + √ 7/25

Since a square root has two values, one positive and the other negative
   x² + (3/5)x - (19/100) = 0
   has two solutions:
  x = -3/10 + √ 7/25
   or
  x = -3/10 - √ 7/25

Note that  √ 7/25 can be written as
  √ 7  / √ 25   which is √ 7  / 5

Solve Quadratic Equation using the Quadratic Formula

 3.3     Solving    100x²+60x-19 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  x  , the solution for   Ax²+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  x =   ————————
                      2A

  In our case,  A   =    100
                      B   =    60
                      C   =  -19

Accordingly,  B²  -  4AC   =
                     3600 - (-7600) =
                     11200

Applying the quadratic formula :

               -60 ± √ 11200
   x  =    ————————
                        200

Can  √ 11200 be simplified ?

Yes!   The prime factorization of  11200   is
   2•2•2•2•2•2•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).

√ 11200   =  √ 2•2•2•2•2•2•5•5•7   =2•2•2•5•√ 7   =
                ±  40 • √ 7

  √ 7   , rounded to 4 decimal digits, is   2.6458
 So now we are looking at:
           x  =  ( -60 ± 40 •  2.646 ) / 200

Two real solutions:

 x =(-60+√11200)/200=-3/10+1/5√ 7 = 0.229

or:

 x =(-60-√11200)/200=-3/10-1/5√ 7 = -0.829

Two solutions were found :

1.  x =(-60-√11200)/200=-3/10-1/5√ 7 = -0.829
2.  x =(-60+√11200)/200=-3/10+1/5√ 7 = 0.229