Question

find roots of2x^2+x+4=0 by quadratic formula

Answer 1

Solving   2x2-x-4 = 0 by Completing The Square .

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

Add  2  to both side of the equation : 
   x2-(1/2)x = 2

Now the clever bit: Take the coefficient of  x , which is  1/2 , divide by two, giving  1/4 , and finally square it giving  1/16 

Add  1/16  to both sides of the equation :
  On the right hand side we have :
   2  +  1/16    or,  (2/1)+(1/16) 
  The common denominator of the two fractions is  16   Adding  (32/16)+(1/16)  gives  33/16 
  So adding to both sides we finally get :
   x2-(1/2)x+(1/16) = 33/16

Adding  1/16  has completed the left hand side into a perfect square :
   x2-(1/2)x+(1/16)  =
   (x-(1/4)) • (x-(1/4))  =
  (x-(1/4))2 
Things which are equal to the same thing are also equal to one another. Since
   x2-(1/2)x+(1/16) = 33/16 and
   x2-(1/2)x+(1/16) = (x-(1/4))2 
then, according to the law of transitivity,
   (x-(1/4))2 = 33/16

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-(1/4))2   is
   (x-(1/4))2/2 =
  (x-(1/4))1 =
   x-(1/4)

Now, applying the Square Root Principle to  Eq. #3.2.1  we get:
   x-(1/4) = √ 33/16 

Add  1/4  to both sides to obtain:
   x = 1/4 + √ 33/16 

Since a square root has two values, one positive and the other negative
   x2 - (1/2)x - 2 = 0
   has two solutions:
  x = 1/4 + √ 33/16 
   or
  x = 1/4 - √ 33/16 

Note that  √ 33/16 can be written as
  √ 33  / √ 16   which is √ 33  / 4 


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