Question

Find the value of root6+root6+root 6.......... Using the completion of square method. Let x = root6+ root6+root6 then square on both the sides. So xsquare=6root6root6root6 . My question is while squaring only one 6 is squared. Why are the other 6 left as roots?

Answer 1

let x =root6 + root6 + root6 + root6 +.......

=root6 {1+ 1+1+1..........n terms}

=root6.n

=n.root6

hence x=n.root6
now squaring both side
x^2=n^2 (root6)^2
=6n^2

x^2-6n^2=0

now use formula ,
a^2-b^2=(a-b)(a+b)

(x^2-6n^2)=(x-nroot6)(x+nroot6)=0
=> x=n.root6 , -nroot6
but x is sum of root6 which is positive so,
x=-n.root6 isn't possible
hence x=n.root6 is only answer