Question

Find the roots of 2Xsaquare _7X+6=0 by the completing square method

Answer 1

Step-by-step explanation:

Divide both sides of the equation by  2  to have 1 as the coefficient of the first term :

   x2-(7/2)x+3 = 0

Subtract  3  from both side of the equation :

   x2-(7/2)x = -3

Now the clever bit: Take the coefficient of  x , which is  7/2 , divide by two, giving  7/4 , and finally square it giving  49/16 

Add  49/16  to both sides of the equation :

  On the right hand side we have :

   -3  +  49/16    or,  (-3/1)+(49/16) 

  The common denominator of the two fractions is  16   Adding  (-48/16)+(49/16)  gives  1/16 

  So adding to both sides we finally get :

   x2-(7/2)x+(49/16) = 1/16

Adding  49/16  has completed the left hand side into a perfect square :

   x2-(7/2)x+(49/16)  =

   (x-(7/4)) • (x-(7/4))  =

  (x-(7/4))2

Things which are equal to the same thing are also equal to one another. Since

   x2-(7/2)x+(49/16) = 1/16 and

   x2-(7/2)x+(49/16) = (x-(7/4))2

then, according to the law of transitivity,

   (x-(7/4))2 = 1/16