Show that $$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots \infty } } } } = 2 $$ when the layers of square root go to infinite

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Show that $$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots \infty  }  }  }  }  =  2 $$ when the layers of square root go to infinite

Uniteasy Admin · Accepted Answer

Let 
$$x = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots \infty } } } }$$c 
when the layers of square root go to infinite
which makes no difference starting to count from the second square root.
Then, we get the following equation
$$x = \sqrt{2+x} $$c
Then,
$$x > 0$$c
Square both sides results in a quadratic equation
$$x^2-x-2=0$$c
Factor the left hand side
$$(x-2)(x+1) = 0$$c
Then,
$$x = 2$$c 
or 
$$x = -1 \:(	ext{cancel})$$c
In summary,
$$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots \infty } } } } = 2$$c

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