Root plot for :Â Â y = x2-4x+15
Axis of Symmetry (dashed)Â Â {x}={ 2.00}Â
Vertex at  {x,y} = { 2.00,11.00} Â
Function has no real roots
Solve Quadratic Equation by Completing The Square
 2.2     Solving   x2-4x+15 = 0 by Completing The Square .
 Subtract  15  from both side of the equation :
   x2-4x = -15
Now the clever bit: Take the coefficient of  x , which is  4 , divide by two, giving  2 , and finally square it giving  4Â
Add  4 to both sides of the equation :
  On the right hand side we have :
   -15  +  4    or, (-15/1)+(4/1)Â
  The common denominator of the two fractions is  1   Adding  (-15/1)+(4/1)  gives  -11/1Â
  So adding to both sides we finally get :
   x2-4x+4 = -11
Adding  4 has completed the left hand side into a perfect square :
   x2-4x+4  =
   (x-2) • (x-2)  =
  (x-2)2Â
Things which are equal to the same thing are also equal to one another. Since
   x2-4x+4 = -11 and
   x2-4x+4 = (x-2)2Â
then, according to the law of transitivity,
   (x-2)2 = -11
We'll refer to this Equation as  Eq. #2.2.1 Â
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
   (x-2)2   is
   (x-2)2/2 =
  (x-2)1 =
   x-2
Now, applying the Square Root Principle to Eq. #2.2.1  we get:
   x-2 = √ -11Â
Add  2  to both sides to obtain:
   x = 2 + √ -11Â
In Math,  i  is called the imaginary unit. It satisfies   i2  =-1. Both   i  and   -i  are the square roots of   -1Â
Since a square root has two values, one positive and the other negative
   x2 - 4x + 15 = 0
   has two solutions:
  x = 2 + √ 11 •  iÂ
   or
  x = 2 - √ 11 •  iÂ
Solve Quadratic Equation using the Quadratic Formula
 2.3     Solving    x2-4x+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 :
                                    Â
            - B  ±  √ B2-4AC
  x =  ————————
                      2AÂ
  In our case,  A   =    1
                      B   =   -4
                      C   =  15Â
Accordingly,  B2  -  4AC   =
                     16 - 60 =
                     -44
Applying the quadratic formula :
               4 ± √ -44Â
   x  =    —————
                    2
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)Â
Both   i  and   -i  are the square roots of minus 1
Accordingly,√ -44  =Â
                    √ 44 • (-1)  =
                    √ 44  • √ -1   =
                    ±  √ 44  • i
Can  √ 44 be simplified ?
Yes!   The prime factorization of  44   is
   2•2•11 Â
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).
√ 44   =  √ 2•2•11   =
                ±  2 • √ 11Â
 √ 11  , rounded to 4 decimal digits, is   3.3166
 So now we are looking at:
           x  =  ( 4 ± 2 • 3.317 i ) / 2
Two imaginary solutions :Â
 x =(4+√-44)/2=2+i√ 11 = 2.0000+3.3166i
  or:
 x =(4-√-44)/2=2-i√ 11 = 2.0000-3.3166i
Two solutions were found : x =(4-√-44)/2=2-i√ 11 = 2.0000-3.3166i x =(4+√-44)/2=2+i√ 11 = 2.0000+3.3166i