Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (31x2 +  260x) +  220  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  31x²+260x+220 

The first term is,  31x²  its coefficient is  31 .
The middle term is,  +260x  its coefficient is  260 .
The last term, "the constant", is  +220 

Step-1 : Multiply the coefficient of the first term by the constant   31 • 220 = 6820 

Step-2 : Find two factors of  6820  whose sum equals the coefficient of the middle term, which is   260 .

For tidiness, printing of 42 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  :

  31x2 + 260x + 220  = 0 

Step  3  :

Parabola, Finding the Vertex :

 3.1      Find the Vertex of   y = 31x²+260x+220

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, 31 , 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  -4.1935  

 Plugging into the parabola formula  -4.1935  for  x  we can calculate the  y -coordinate : 
  y = 31.0 * -4.19 * -4.19 + 260.0 * -4.19 + 220.0
or   y = -325.161

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 31x²+260x+220
Axis of Symmetry (dashed)  {x}={-4.19} 
Vertex at  {x,y} = {-4.19,-325.16} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-7.43, 0.00} 
Root 2 at  {x,y} = {-0.95, 0.00} 

Solve Quadratic Equation by Completing The Square

 3.2     Solving   31x²+260x+220 = 0 by Completing The Square .

 Divide both sides of the equation by  31  to have 1 as the coefficient of the first term :
   x²+(260/31)x+(220/31) = 0

Subtract  220/31  from both side of the equation :
   x²+(260/31)x = -220/31

Now the clever bit: Take the coefficient of  x , which is  260/31 , divide by two, giving  130/31 , and finally square it giving  16900/961 

Add  16900/961  to both sides of the equation :
  On the right hand side we have :
   -220/31  +  16900/961   The common denominator of the two fractions is  961   Adding  (-6820/961)+(16900/961)  gives  10080/961 
  So adding to both sides we finally get :
   x²+(260/31)x+(16900/961) = 10080/961

Adding  16900/961  has completed the left hand side into a perfect square :
   x²+(260/31)x+(16900/961)  =
   (x+(130/31)) • (x+(130/31))  =
  (x+(130/31))²
Things which are equal to the same thing are also equal to one another. Since
   x²+(260/31)x+(16900/961) = 10080/961 and
   x²+(260/31)x+(16900/961) = (x+(130/31))²
then, according to the law of transitivity,
   (x+(130/31))² = 10080/961

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+(130/31))²   is
   (x+(130/31))^2/2 =
  (x+(130/31))¹ =
   x+(130/31)

Now, applying the Square Root Principle to  Eq. #3.2.1  we get:
   x+(130/31) = √ 10080/961

Subtract  130/31  from both sides to obtain:
   x = -130/31 + √ 10080/961

Since a square root has two values, one positive and the other negative
   x² + (260/31)x + (220/31) = 0
   has two solutions:
  x = -130/31 + √ 10080/961
   or
  x = -130/31 - √ 10080/961

Note that  √ 10080/961 can be written as
  √ 10080  / √ 961   which is √ 10080  / 31

Solve Quadratic Equation using the Quadratic Formula

 3.3     Solving    31x²+260x+220 = 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   =     31
                      B   =   260
                      C   =  220

Accordingly,  B²  -  4AC   =
                     67600 - 27280 =
                     40320

Applying the quadratic formula :

               -260 ± √ 40320
   x  =    ————————
                        62

Can  √ 40320 be simplified ?

Yes!   The prime factorization of  40320   is
   2•2•2•2•2•2•2•3•3•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).

√ 40320   =  √ 2•2•2•2•2•2•2•3•3•5•7   =2•2•2•3•√ 70   =
                ±  24 • √ 70

  √ 70   , rounded to 4 decimal digits, is   8.3666
 So now we are looking at:
           x  =  ( -260 ± 24 •  8.367 ) / 62

Two real solutions:

 x =(-260+√40320)/62=(-130+12√ 70 )/31= -0.955

or:

 x =(-260-√40320)/62=(-130-12√ 70 )/31= -7.432

Two solutions were found :

1.  x =(-260-√40320)/62=(-130-12√ 70 )/31= -7.432
2.  x =(-260+√40320)/62=(-130+12√ 70 )/31= -0.955