Question

find the value of polynomial p(y) = 6y^2-7y+√3 at y =√3

Answer 1

Heya !!!



Y = root 3




P(X) = 6Y² - 7Y + root 3



P(root 3) = 6 × ( root 3)² - 7 × root 3 + root 3



=> 6 × 3 - 7root 3 + root 3




=> 18 - 6 root 3



=> 6 ( 3 - root 3)





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Answer 2

Divide both sides of the equation by  6  to have 1 as the coefficient of the first term :
   y2-(7/6)y-(1/2) = 0

Add  1/2  to both side of the equation : 
   y2-(7/6)y = 1/2

Now the clever bit: Take the coefficient of  y , which is  7/6 , divide by two, giving  7/12 , and finally square it giving  49/144 

Add  49/144  to both sides of the equation :
  On the right hand side we have :
   1/2  +  49/144   The common denominator of the two fractions is  144   Adding  (72/144)+(49/144)  gives  121/144 
  So adding to both sides we finally get :
   y2-(7/6)y+(49/144) = 121/144

Adding  49/144  has completed the left hand side into a perfect square :
   y2-(7/6)y+(49/144)  =
   (y-(7/12)) • (y-(7/12))  =
  (y-(7/12))2 
Things which are equal to the same thing are also equal to one another. Since
   y2-(7/6)y+(49/144) = 121/144 and
   y2-(7/6)y+(49/144) = (y-(7/12))2 
then, according to the law of transitivity,
   (y-(7/12))2 = 121/144

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
   (y-(7/12))2   is
   (y-(7/12))2/2 =
  (y-(7/12))1 =
   y-(7/12)

Now, applying the Square Root Principle to  Eq. #4.2.1  we get:
   y-(7/12) = √ 121/144 

Add  7/12  to both sides to obtain:
   y = 7/12 + √ 121/144 

Since a square root has two values, one positive and the other negative
   y2 - (7/6)y - (1/2) = 0
   has two solutions:
  y = 7/12 + √ 121/144 
   or
  y = 7/12 - √ 121/144 

Note that  √ 121/144 can be written as
  √ 121  / √ 144   which is 11 / 12