Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((3•19x2) +  118x) -  3  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  57x²+118x-3 

The first term is,  57x²  its coefficient is  57 .
The middle term is,  +118x  its coefficient is  118 .
The last term, "the constant", is  -3 

Step-1 : Multiply the coefficient of the first term by the constant   57 • -3 = -171 

Step-2 : Find two factors of  -171  whose sum equals the coefficient of the middle term, which is   118 .

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

Equation at the end of step  2  :

  57x2 + 118x - 3  = 0 

Step  3  :

Parabola, Finding the Vertex :

 3.1      Find the Vertex of   y = 57x²+118x-3

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, 57 , 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  -1.0351  

 Plugging into the parabola formula  -1.0351  for  x  we can calculate the  y -coordinate : 
  y = 57.0 * -1.04 * -1.04 + 118.0 * -1.04 - 3.0
or   y = -64.070

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 57x²+118x-3
Axis of Symmetry (dashed)  {x}={-1.04} 
Vertex at  {x,y} = {-1.04,-64.07} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-2.10, 0.00} 
Root 2 at  {x,y} = { 0.03, 0.00} 

Solve Quadratic Equation by Completing The Square

 3.2     Solving   57x²+118x-3 = 0 by Completing The Square .

 Divide both sides of the equation by  57  to have 1 as the coefficient of the first term :
   x²+(118/57)x-(1/19) = 0

Add  1/19  to both side of the equation :
   x²+(118/57)x = 1/19

Now the clever bit: Take the coefficient of  x , which is  118/57 , divide by two, giving  59/57 , and finally square it giving  3481/3249 

Add  3481/3249  to both sides of the equation :
  On the right hand side we have :
   1/19  +  3481/3249   The common denominator of the two fractions is  3249   Adding  (171/3249)+(3481/3249)  gives  3652/3249 
  So adding to both sides we finally get :
   x²+(118/57)x+(3481/3249) = 3652/3249

Adding  3481/3249  has completed the left hand side into a perfect square :
   x²+(118/57)x+(3481/3249)  =
   (x+(59/57)) • (x+(59/57))  =
  (x+(59/57))²
Things which are equal to the same thing are also equal to one another. Since
   x²+(118/57)x+(3481/3249) = 3652/3249 and
   x²+(118/57)x+(3481/3249) = (x+(59/57))²
then, according to the law of transitivity,
   (x+(59/57))² = 3652/3249

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+(59/57))²   is
   (x+(59/57))^2/2 =
  (x+(59/57))¹ =
   x+(59/57)

Now, applying the Square Root Principle to  Eq. #3.2.1  we get:
   x+(59/57) = √ 3652/3249

Subtract  59/57  from both sides to obtain:
   x = -59/57 + √ 3652/3249

Since a square root has two values, one positive and the other negative
   x² + (118/57)x - (1/19) = 0
   has two solutions:
  x = -59/57 + √ 3652/3249
   or
  x = -59/57 - √ 3652/3249

Note that  √ 3652/3249 can be written as
  √ 3652  / √ 3249   which is √ 3652  / 57

Solve Quadratic Equation using the Quadratic Formula

 3.3     Solving    57x²+118x-3 = 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   =     57
                      B   =   118
                      C   =   -3

Accordingly,  B²  -  4AC   =
                     13924 - (-684) =
                     14608

Applying the quadratic formula :

               -118 ± √ 14608
   x  =    ————————
                        114

Can  √ 14608 be simplified ?

Yes!   The prime factorization of  14608   is
   2•2•2•2•11•83 
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).

√ 14608   =  √ 2•2•2•2•11•83   =2•2•√ 913   =
                ±  4 • √ 913

  √ 913   , rounded to 4 decimal digits, is  30.2159
 So now we are looking at:
           x  =  ( -118 ± 4 •  30.216 ) / 114

Two real solutions:

 x =(-118+√14608)/114=(-59+2√ 913 )/57= 0.025

or:

 x =(-118-√14608)/114=(-59-2√ 913 )/57= -2.095

Two solutions were found :

1.  x =(-118-√14608)/114=(-59-2√ 913 )/57= -2.095
2.  x =(-118+√14608)/114=(-59+2√ 913 )/57= 0.025