Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((2•5•101x2) +  626x) +  81  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  1010x²+626x+81 

The first term is,  1010x²  its coefficient is  1010 .
The middle term is,  +626x  its coefficient is  626 .
The last term, "the constant", is  +81 

Step-1 : Multiply the coefficient of the first term by the constant   1010 • 81 = 81810 

Step-2 : Find two factors of  81810  whose sum equals the coefficient of the middle term, which is   626 .

For tidiness, printing of 74 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  :

  1010x2 + 626x + 81  = 0 

Step  3  :

Parabola, Finding the Vertex :

 3.1      Find the Vertex of   y = 1010x²+626x+81

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, 1010 , 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  -0.3099  

 Plugging into the parabola formula  -0.3099  for  x  we can calculate the  y -coordinate : 
  y = 1010.0 * -0.31 * -0.31 + 626.0 * -0.31 + 81.0
or   y = -15.999

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 1010x²+626x+81
Axis of Symmetry (dashed)  {x}={-0.31} 
Vertex at  {x,y} = {-0.31,-16.00} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-0.44, 0.00} 
Root 2 at  {x,y} = {-0.18, 0.00} 

Solve Quadratic Equation by Completing The Square

 3.2     Solving   1010x²+626x+81 = 0 by Completing The Square .

 Divide both sides of the equation by  1010  to have 1 as the coefficient of the first term :
   x²+(313/505)x+(81/1010) = 0

Subtract  81/1010  from both side of the equation :
   x²+(313/505)x = -81/1010

Now the clever bit: Take the coefficient of  x , which is  313/505 , divide by two, giving  313/1010 , and finally square it giving  313/1010 

Add  313/1010  to both sides of the equation :
  On the right hand side we have :
   -81/1010  +  313/1010   The common denominator of the two fractions is  1010   Adding  (-81/1010)+(313/1010)  gives  232/1010 
  So adding to both sides we finally get :
   x²+(313/505)x+(313/1010) = 116/505

Adding  313/1010  has completed the left hand side into a perfect square :
   x²+(313/505)x+(313/1010)  =
   (x+(313/1010)) • (x+(313/1010))  =
  (x+(313/1010))²
Things which are equal to the same thing are also equal to one another. Since
   x²+(313/505)x+(313/1010) = 116/505 and
   x²+(313/505)x+(313/1010) = (x+(313/1010))²
then, according to the law of transitivity,
   (x+(313/1010))² = 116/505

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+(313/1010))²   is
   (x+(313/1010))^2/2 =
  (x+(313/1010))¹ =
   x+(313/1010)

Now, applying the Square Root Principle to  Eq. #3.2.1  we get:
   x+(313/1010) = √ 116/505

Subtract  313/1010  from both sides to obtain:
   x = -313/1010 + √ 116/505

Since a square root has two values, one positive and the other negative
   x² + (313/505)x + (81/1010) = 0
   has two solutions:
  x = -313/1010 + √ 116/505
   or
  x = -313/1010 - √ 116/505

Note that  √ 116/505 can be written as
  √ 116  / √ 505 

Solve Quadratic Equation using the Quadratic Formula

 3.3     Solving    1010x²+626x+81 = 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   =    1010
                      B   =   626
                      C   =   81

Accordingly,  B²  -  4AC   =
                     391876 - 327240 =
                     64636

Applying the quadratic formula :

               -626 ± √ 64636
   x  =    ————————
                        2020

Can  √ 64636 be simplified ?

Yes!   The prime factorization of  64636   is
   2•2•11•13•113 
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).

√ 64636   =  √ 2•2•11•13•113   =
                ±  2 • √ 16159

  √ 16159   , rounded to 4 decimal digits, is  127.1181
 So now we are looking at:
           x  =  ( -626 ± 2 •  127.118 ) / 2020

Two real solutions:

 x =(-626+√64636)/2020=(-313+√ 16159 )/1010= -0.184

or:

 x =(-626-√64636)/2020=(-313-√ 16159 )/1010= -0.436

Two solutions were found :

1.  x =(-626-√64636)/2020=(-313-√ 16159 )/1010= -0.436
2.  x =(-626+√64636)/2020=(-313+√ 16159 )/1010= -0.184