# Thread: Problem based on square root

1. ## Problem based on square root

If $\displaystyle \sqrt{6+\sqrt{6+\sqrt{6+......}}}$, then

a) x=3
b) x= $\displaystyle \infty$
c) x=6
d) x=9

2. Originally Posted by siddscool19
If $\displaystyle \sqrt{6+\sqrt{6+\sqrt{6+......}}}$, then

a) x=3
b) x= $\displaystyle \infty$
c) x=6
d) x=9

The answer is a. As you take the square root of 6 infinitely and continuously add it to itself, you will slowly approach 1/2 of 6 which is three. You will get closer and closer to three, but the repeating square root prohibits is from getting there.

3. Let: $\displaystyle x = \sqrt{6+\sqrt{6+\sqrt{6+ \cdots}}} > 0$

Now, note that if you square it and move +6 to the LHS: $\displaystyle x^2 = 6 + \sqrt{6+\sqrt{6+ \cdots}} \ \ \Leftrightarrow \ \ x^2 - 6 = \sqrt{6+\sqrt{6+ \cdots}}$

But the RHS is $\displaystyle x$ again!

Thus, we see that $\displaystyle x$ is the positive solution to the quadratic: $\displaystyle x^2 - x - 6 = 0$

which is simply 3.

4. Originally Posted by o_O
Let: $\displaystyle x = \sqrt{6+\sqrt{6+\sqrt{6+ \cdots}}} > 0$

Now, note that if you square it and move +6 to the LHS: $\displaystyle x^2 = 6 + \sqrt{6+\sqrt{6+ \cdots}} \ \ \Leftrightarrow \ \ x^2 - 6 = \sqrt{6+\sqrt{6+ \cdots}}$

But the RHS is $\displaystyle x$ again!

Thus, we see that $\displaystyle x$ is the positive solution to the quadratic: $\displaystyle x^2 - x - 6 = 0$

which is simply 3.
$\displaystyle x = \sqrt{6+\sqrt{6+\sqrt{6+ \cdots}}} > 0$
$\displaystyle x=\sqrt{6+x}$
$\displaystyle x^2=6+x$
x=3 or -2

ans=(a)3