Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (22x2 +  124x) -  520  = 0 

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   4x² + 124x - 520  =   4 • (x² + 31x - 130) 

Trying to factor by splitting the middle term

 3.2     Factoring  x² + 31x - 130 

The first term is,  x²  its coefficient is  1 .
The middle term is,  +31x  its coefficient is  31 .
The last term, "the constant", is  -130 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -130 = -130 

Step-2 : Find two factors of  -130  whose sum equals the coefficient of the middle term, which is   31 .

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

Equation at the end of step  3  :

  4 • (x2 + 31x - 130)  = 0 

Step  4  :

Equations which are never true :

 4.1      Solve :    4   =  0

This equation has no solution.
A a non-zero constant never equals zero.

Parabola, Finding the Vertex :

 4.2      Find the Vertex of   y = x²+31x-130

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, 1 , 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  -15.5000  

 Plugging into the parabola formula  -15.5000  for  x  we can calculate the  y -coordinate : 
  y = 1.0 * -15.50 * -15.50 + 31.0 * -15.50 - 130.0
or   y = -370.250

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x²+31x-130
Axis of Symmetry (dashed)  {x}={-15.50} 
Vertex at  {x,y} = {-15.50,-370.25} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-34.74, 0.00} 
Root 2 at  {x,y} = { 3.74, 0.00} 

Solve Quadratic Equation by Completing The Square

 4.3     Solving   x²+31x-130 = 0 by Completing The Square .

 Add  130  to both side of the equation :
   x²+31x = 130

Now the clever bit: Take the coefficient of  x , which is  31 , divide by two, giving  31/2 , and finally square it giving  961/4 

Add  961/4  to both sides of the equation :
  On the right hand side we have :
   130  +  961/4    or,  (130/1)+(961/4) 
  The common denominator of the two fractions is  4   Adding  (520/4)+(961/4)  gives  1481/4 
  So adding to both sides we finally get :
   x²+31x+(961/4) = 1481/4

Adding  961/4  has completed the left hand side into a perfect square :
   x²+31x+(961/4)  =
   (x+(31/2)) • (x+(31/2))  =
  (x+(31/2))²
Things which are equal to the same thing are also equal to one another. Since
   x²+31x+(961/4) = 1481/4 and
   x²+31x+(961/4) = (x+(31/2))²
then, according to the law of transitivity,
   (x+(31/2))² = 1481/4

We'll refer to this Equation as  Eq. #4.3.1  

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

Note that the square root of
   (x+(31/2))²   is
   (x+(31/2))^2/2 =
  (x+(31/2))¹ =
   x+(31/2)

Now, applying the Square Root Principle to  Eq. #4.3.1  we get:
   x+(31/2) = √ 1481/4

Subtract  31/2  from both sides to obtain:
   x = -31/2 + √ 1481/4

Since a square root has two values, one positive and the other negative
   x² + 31x - 130 = 0
   has two solutions:
  x = -31/2 + √ 1481/4
   or
  x = -31/2 - √ 1481/4

Note that  √ 1481/4 can be written as
  √ 1481  / √ 4   which is √ 1481  / 2

Solve Quadratic Equation using the Quadratic Formula

 4.4     Solving    x²+31x-130 = 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   =     1
                      B   =    31
                      C   =  -130

Accordingly,  B²  -  4AC   =
                     961 - (-520) =
                     1481

Applying the quadratic formula :

               -31 ± √ 1481
   x  =    ———————
                        2

  √ 1481   , rounded to 4 decimal digits, is  38.4838
 So now we are looking at:
           x  =  ( -31 ±  38.484 ) / 2

Two real solutions:

 x =(-31+√1481)/2= 3.742

or:

 x =(-31-√1481)/2=-34.742

Two solutions were found :

1.  x =(-31-√1481)/2=-34.742
2.  x =(-31+√1481)/2= 3.742