Question

Devise the plan to find the value of x.

Answer 1

the square roots of 2 continue forever

Answer 2

check if my way is correct
\[x=\sqrt{2+\sqrt{2+\sqrt{2+...}}}\]
\[x^2=2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}\]
thus
\[x^2-2=x\]
solving yields
x=-1
and x=2
but x can't be -1 because a principal square root can't be negative
thus x is 2.
\[\sqrt{2+\sqrt{2+\sqrt{2+...}}}=2\]

Answer 3

thx guys :)

Answer 4

yes, that's correct