Solution - Quadratic equations

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((2•5x2) +  31x) -  15  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  10x²+31x-15 

The first term is,  10x²  its coefficient is  10 .
The middle term is,  +31x  its coefficient is  31 .
The last term, "the constant", is  -15 

Step-1 : Multiply the coefficient of the first term by the constant   10 • -15 = -150 

Step-2 : Find two factors of  -150  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  2  :

  10x2 + 31x - 15  = 0 

Step  3  :

Parabola, Finding the Vertex :

 3.1      Find the Vertex of   y = 10x²+31x-15

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, 10 , 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  -1.5500  

 Plugging into the parabola formula  -1.5500  for  x  we can calculate the  y -coordinate : 
  y = 10.0 * -1.55 * -1.55 + 31.0 * -1.55 - 15.0
or   y = -39.025

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 10x²+31x-15
Axis of Symmetry (dashed)  {x}={-1.55} 
Vertex at  {x,y} = {-1.55,-39.03} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-3.53, 0.00} 
Root 2 at  {x,y} = { 0.43, 0.00} 

Solve Quadratic Equation by Completing The Square

 3.2     Solving   10x²+31x-15 = 0 by Completing The Square .

 Divide both sides of the equation by  10  to have 1 as the coefficient of the first term :
   x²+(31/10)x-(3/2) = 0

Add  3/2  to both side of the equation :
   x²+(31/10)x = 3/2

Now the clever bit: Take the coefficient of  x , which is  31/10 , divide by two, giving  31/20 , and finally square it giving  961/400 

Add  961/400  to both sides of the equation :
  On the right hand side we have :
   3/2  +  961/400   The common denominator of the two fractions is  400   Adding  (600/400)+(961/400)  gives  1561/400 
  So adding to both sides we finally get :
   x²+(31/10)x+(961/400) = 1561/400

Adding  961/400  has completed the left hand side into a perfect square :
   x²+(31/10)x+(961/400)  =
   (x+(31/20)) • (x+(31/20))  =
  (x+(31/20))²
Things which are equal to the same thing are also equal to one another. Since
   x²+(31/10)x+(961/400) = 1561/400 and
   x²+(31/10)x+(961/400) = (x+(31/20))²
then, according to the law of transitivity,
   (x+(31/20))² = 1561/400

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+(31/20))²   is
   (x+(31/20))^2/2 =
  (x+(31/20))¹ =
   x+(31/20)

Now, applying the Square Root Principle to  Eq. #3.2.1  we get:
   x+(31/20) = √ 1561/400

Subtract  31/20  from both sides to obtain:
   x = -31/20 + √ 1561/400

Since a square root has two values, one positive and the other negative
   x² + (31/10)x - (3/2) = 0
   has two solutions:
  x = -31/20 + √ 1561/400
   or
  x = -31/20 - √ 1561/400

Note that  √ 1561/400 can be written as
  √ 1561  / √ 400   which is √ 1561  / 20

Solve Quadratic Equation using the Quadratic Formula

 3.3     Solving    10x²+31x-15 = 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   =     10
                      B   =    31
                      C   =  -15

Accordingly,  B²  -  4AC   =
                     961 - (-600) =
                     1561

Applying the quadratic formula :

               -31 ± √ 1561
   x  =    ———————
                        20

  √ 1561   , rounded to 4 decimal digits, is  39.5095
 So now we are looking at:
           x  =  ( -31 ±  39.509 ) / 20

Two real solutions:

 x =(-31+√1561)/20= 0.425

or:

 x =(-31-√1561)/20=-3.525

Two solutions were found :

1.  x =(-31-√1561)/20=-3.525
2.  x =(-31+√1561)/20= 0.425