Solution - Quadratic equations

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                     4*x-((15*x^2+5*x+14))=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  4x -  (((3•5x2) +  5x) +  14)  = 0 

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   -15x² - x - 14  =   -1 • (15x² + x + 14) 

Trying to factor by splitting the middle term

 3.2     Factoring  15x² + x + 14 

The first term is,  15x²  its coefficient is  15 .
The middle term is,  +x  its coefficient is  1 .
The last term, "the constant", is  +14 

Step-1 : Multiply the coefficient of the first term by the constant   15 • 14 = 210 

Step-2 : Find two factors of  210  whose sum equals the coefficient of the middle term, which is   1 .

For tidiness, printing of 26 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  3  :

  -15x2 - x - 14  = 0 

Step  4  :

Parabola, Finding the Vertex :

 4.1      Find the Vertex of   y = -15x²-x-14

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens down and accordingly has a highest point (AKA absolute maximum) .    We know this even before plotting  "y"  because the coefficient of the first term, -15 , is negative (smaller 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  -0.0333  

 Plugging into the parabola formula  -0.0333  for  x  we can calculate the  y -coordinate : 
  y = -15.0 * -0.03 * -0.03 - 1.0 * -0.03 - 14.0
or   y = -13.983

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = -15x²-x-14
Axis of Symmetry (dashed)  {x}={-0.03} 
Vertex at  {x,y} = {-0.03,-13.98} 
Function has no real roots

Solve Quadratic Equation by Completing The Square

 4.2     Solving   -15x²-x-14 = 0 by Completing The Square .

 Multiply both sides of the equation by  (-1)  to obtain positive coefficient for the first term:
 15x²+x+14 = 0  Divide both sides of the equation by  15  to have 1 as the coefficient of the first term :
   x²+(1/15)x+(14/15) = 0

Subtract  14/15  from both side of the equation :
   x²+(1/15)x = -14/15

Now the clever bit: Take the coefficient of  x , which is  1/15 , divide by two, giving  1/30 , and finally square it giving  1/900 

Add  1/900  to both sides of the equation :
  On the right hand side we have :
   -14/15  +  1/900   The common denominator of the two fractions is  900   Adding  (-840/900)+(1/900)  gives  -839/900 
  So adding to both sides we finally get :
   x²+(1/15)x+(1/900) = -839/900

Adding  1/900  has completed the left hand side into a perfect square :
   x²+(1/15)x+(1/900)  =
   (x+(1/30)) • (x+(1/30))  =
  (x+(1/30))²
Things which are equal to the same thing are also equal to one another. Since
   x²+(1/15)x+(1/900) = -839/900 and
   x²+(1/15)x+(1/900) = (x+(1/30))²
then, according to the law of transitivity,
   (x+(1/30))² = -839/900

We'll refer to this Equation as  Eq. #4.2.1  

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

Note that the square root of
   (x+(1/30))²   is
   (x+(1/30))^2/2 =
  (x+(1/30))¹ =
   x+(1/30)

Now, applying the Square Root Principle to  Eq. #4.2.1  we get:
   x+(1/30) = √ -839/900

Subtract  1/30  from both sides to obtain:
   x = -1/30 + √ -839/900
In Math,  i  is called the imaginary unit. It satisfies   i²  =-1. Both   i   and   -i   are the square roots of   -1 


Since a square root has two values, one positive and the other negative
   x² + (1/15)x + (14/15) = 0
   has two solutions:
  x = -1/30 + √ 839/900 •  i 
   or
  x = -1/30 - √ 839/900 •  i 

Note that  √ 839/900 can be written as
  √ 839  / √ 900   which is √ 839  / 30

Solve Quadratic Equation using the Quadratic Formula

 4.3     Solving    -15x²-x-14 = 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   =    -15
                      B   =    -1
                      C   =  -14

Accordingly,  B²  -  4AC   =
                     1 - 840 =
                     -839

Applying the quadratic formula :

               1 ± √ -839
   x  =    ——————
                      -30

In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written  (a+b*i) 

Both   i   and   -i   are the square roots of minus 1

Accordingly,√ -839  = 
                    √ 839 • (-1)  =
                    √ 839  • √ -1   =
                    ±  √ 839  • i

  √ 839   , rounded to 4 decimal digits, is  28.9655
 So now we are looking at:
           x  =  ( 1 ±  28.965 i ) / -30

Two imaginary solutions :

 x =(1+√-839)/-30=1/-30-i/30√ 839 = -0.0333+0.9655i
  or: 
 x =(1-√-839)/-30=1/-30+i/30√ 839 = -0.0333-0.9655i

Two solutions were found :

1.  x =(1-√-839)/-30=1/-30+i/30√ 839 = -0.0333-0.9655i
2.  x =(1+√-839)/-30=1/-30-i/30√ 839 = -0.0333+0.9655i