Solution - Quadratic equations

Rearrange:

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

                     0-(6*x^2+45*x+50)=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  0 -  (((2•3x2) +  45x) +  50)  = 0 

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   -6x² - 45x - 50  =   -1 • (6x² + 45x + 50) 

Trying to factor by splitting the middle term

 3.2     Factoring  6x² + 45x + 50 

The first term is,  6x²  its coefficient is  6 .
The middle term is,  +45x  its coefficient is  45 .
The last term, "the constant", is  +50 

Step-1 : Multiply the coefficient of the first term by the constant   6 • 50 = 300 

Step-2 : Find two factors of  300  whose sum equals the coefficient of the middle term, which is   45 .

For tidiness, printing of 30 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  :

  -6x2 - 45x - 50  = 0 

Step  4  :

Parabola, Finding the Vertex :

 4.1      Find the Vertex of   y = -6x²-45x-50

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens down and accordingly has a highest point (AKA absolute maximum) .    We know this even before plotting  "y"  because the coefficient of the first term, -6 , is negative (smaller 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  -3.7500  

 Plugging into the parabola formula  -3.7500  for  x  we can calculate the  y -coordinate : 
  y = -6.0 * -3.75 * -3.75 - 45.0 * -3.75 - 50.0
or   y = 34.375

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = -6x²-45x-50
Axis of Symmetry (dashed)  {x}={-3.75} 
Vertex at  {x,y} = {-3.75,34.38} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-1.36, 0.00} 
Root 2 at  {x,y} = {-6.14, 0.00} 

Solve Quadratic Equation by Completing The Square

 4.2     Solving   -6x²-45x-50 = 0 by Completing The Square .

 Multiply both sides of the equation by  (-1)  to obtain positive coefficient for the first term:
 6x²+45x+50 = 0  Divide both sides of the equation by  6  to have 1 as the coefficient of the first term :
   x²+(15/2)x+(25/3) = 0

Subtract  25/3  from both side of the equation :
   x²+(15/2)x = -25/3

Now the clever bit: Take the coefficient of  x , which is  15/2 , divide by two, giving  15/4 , and finally square it giving  225/16 

Add  225/16  to both sides of the equation :
  On the right hand side we have :
   -25/3  +  225/16   The common denominator of the two fractions is  48   Adding  (-400/48)+(675/48)  gives  275/48 
  So adding to both sides we finally get :
   x²+(15/2)x+(225/16) = 275/48

Adding  225/16  has completed the left hand side into a perfect square :
   x²+(15/2)x+(225/16)  =
   (x+(15/4)) • (x+(15/4))  =
  (x+(15/4))²
Things which are equal to the same thing are also equal to one another. Since
   x²+(15/2)x+(225/16) = 275/48 and
   x²+(15/2)x+(225/16) = (x+(15/4))²
then, according to the law of transitivity,
   (x+(15/4))² = 275/48

We'll refer to this Equation as  Eq. #4.2.1  

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

Note that the square root of
   (x+(15/4))²   is
   (x+(15/4))^2/2 =
  (x+(15/4))¹ =
   x+(15/4)

Now, applying the Square Root Principle to  Eq. #4.2.1  we get:
   x+(15/4) = √ 275/48

Subtract  15/4  from both sides to obtain:
   x = -15/4 + √ 275/48

Since a square root has two values, one positive and the other negative
   x² + (15/2)x + (25/3) = 0
   has two solutions:
  x = -15/4 + √ 275/48
   or
  x = -15/4 - √ 275/48

Note that  √ 275/48 can be written as
  √ 275  / √ 48 

Solve Quadratic Equation using the Quadratic Formula

 4.3     Solving    -6x²-45x-50 = 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   =     -6
                      B   =   -45
                      C   =  -50

Accordingly,  B²  -  4AC   =
                     2025 - 1200 =
                     825

Applying the quadratic formula :

               45 ± √ 825
   x  =    ——————
                      -12

Can  √ 825 be simplified ?

Yes!   The prime factorization of  825   is
   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).

√ 825   =  √ 3•5•5•11   =
                ±  5 • √ 33

  √ 33   , rounded to 4 decimal digits, is   5.7446
 So now we are looking at:
           x  =  ( 45 ± 5 •  5.745 ) / -12

Two real solutions:

 x =(45+√825)/-12=15/-4-5/12√ 33 = -6.144

or:

 x =(45-√825)/-12=15/-4+5/12√ 33 = -1.356

Two solutions were found :

1.  x =(45-√825)/-12=15/-4+5/12√ 33 = -1.356
2.  x =(45+√825)/-12=15/-4-5/12√ 33 = -6.144