Square roots of square roots of

$x=\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}$

$x=\sqrt{6+x}\Rightarrow x^2=6+x$

$x^2-x-6=0$

$(x-3)(x+2)=0$

Because the answer must be positive, the only solution we're interested in is $x = \boxed{3}.$

This question is not as straightforward to do properly as one might think.

All of the posted solutions are only valid once you know that the expression converges. For example, consider:

$x = 1 + 2 + 4 + 8 + 16 + .....$

Obviously this series does not converge; however, using the method given in the other solutions:

$x = 1 + 2 + 4 + 8 + 16 + .....$ $= 1 + 2(1 + 2 + 4 + 8 + 16 + .....)$ $= 1 + 2 x$

$\Rightarrow x = -1$ , which is obviously not correct.

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So first we need to prove that $x=\sqrt{6+\sqrt{6+\sqrt{6+\dots}}}$ converges. What's meant by the expression is also not completely clear in a rigorous sense, so I'll define it as follows:

Suppose $x_0=\sqrt{6}$ , and let $x_i=\sqrt{6+x_{i-1}}$ , for $i=1, 2, \dots$ . Then we define

$x = \sqrt{6+\sqrt{6+\sqrt{6+\dots}}} =\lim_{i\rightarrow \infty} x_i$

We want to show that this sequence is bounded and monotonic: in this case, we want to show that it is bounded above, and increasing. Then we will know that the limit exists and is finite (and that $\lim_{i\rightarrow\infty} x_i = \sup\{x_i:i=0,1,\dots\}$ ).

Let's look at the region in which the sequence is increasing, i.e. $x_{i+1} > x_i$ .

$x_{i+1}>x_i$ $\Rightarrow \sqrt{6+x_i} > x_i$ $\Rightarrow x_i^2-x_i-6<0$ $\Rightarrow -2 < x_i < 3$

For $x_i>-2$ , we see that $x_{i+1}=\sqrt{6+x_i}>\sqrt{4}=2>-2$ . For $x_i<3$ , we see that $x_{i+1}=\sqrt{6+x_i}<\sqrt{9}=3$ . Thus if $x_i$ is in an increasing region, so is $x_{i+1}$ . Since $-2<x_0<3$ , by induction we see that $-2<x_i<3\hspace{0.5em} \forall i$ .

So we have bounded the sequence above by the value $3$ , and checked that the sequence is increasing. Thus the limit exists and is finite. Thus we can solve $x_{i+1}=x_i$ to get the limit. Since the sequence is increasing, we of course want the higher of these solutions, $x=3$ .

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While my solution might seem like overkill, hopefully the first example I gave should show why it is necessary to prove that the limit exists and is finite before trying to find it!

using the generalization sqrt(n+sqrt(n+sqrt(n+...)=s

s^2=n+sqrt(n+sqrt(n+...)

s^2=n+s

s^2-s=n

s(s-1)=n

in this case n=6

s(s-1)=6

s=3

I saw that √6 is 2.49–one of the answers–and knew that it would be a bit more than that, but not quite to 4, the square root of 16. The closest one had to be 3.

Why does x=√(6+x) ? I dont understand this, cuz √(6+x) will be a different number or something, so it cant be x.

Assuming as variable y => y = (6 + y) ^ 1/2> 6 + y> y^2 - 6 -2 = 0 >> y = 6 / -3 and y = 6/2; conclusion: a = 3. Given that y = -2 assumes the form of actual number in a square root, the answer is> 3. S = {3, -2}

x= √(6 + √(6 + √(6 + .....)))

x=√(6 + x)

x^2= 6+x

x^2-x-6=0

x=3 (positive value)