Endless Rooting

$\sqrt{5 \sqrt{5\sqrt{5........}}}$

0r, $\sqrt{5x}=x$

Or, $5x=x^{2}$

Or, $x=5$

$\sqrt{5\sqrt{5\sqrt{5\sqrt{5...}}}}=?$

$\Rightarrow 5\sqrt{5\sqrt{5\sqrt{5...}}}=?^2$

$\Rightarrow 5?=?^2$

$\Rightarrow \boxed{5=?}$

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Or,

$5?=?^2$

$\Rightarrow 5?-?^2=0$

$\Rightarrow ?(5-?)=0$

$\Rightarrow ?=0,5$

Since $?\neq 0$ ,

$\boxed{?=5}$ .

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(In math, it is generally more proper to solve by factoring rather than canceling like in the first solution. This is because canceling sometimes leads to loss of roots. And even though the root lost in this case was a false one, sometimes it may be a true one.)

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GENERAL CASE:

$\boxed{\sqrt{N\sqrt{N\sqrt{N\sqrt{N...}}}}=N}$

Proof:

$\sqrt{N\sqrt{N\sqrt{N\sqrt{N...}}}}=N$

$\Rightarrow N\sqrt{N\sqrt{N\sqrt{N...}}}=N^2$

$\Rightarrow N\cdot N=N^2$

$\Rightarrow \boxed{N^2=N^2}$ ,

$\forall N$ .

Let the expression be equal to $x$ . Then squaring both sides gives $x^2=5x$ . If you don't see why, notice how by squaring $x$ adds a coefficient of $5$ to $x$ . Solving, we have $x=5$ .

Let $\sqrt { 5\sqrt { 5\sqrt { 5.... } } } =x$

Then ${ x }^{ 2 }=5\sqrt { 5\sqrt { 5\sqrt { 5.... } } }$

We already know that $\sqrt { 5\sqrt { 5\sqrt { 5.... } } } =x$ , so $5\sqrt { 5\sqrt { 5\sqrt { 5.... } } }=5x$

It becomes ${ x }^{ 2 }=5x$ and $x=0\quad and\quad 5$

So $\boxed{x=5}$

If X equal the above equation. Then x^2 will equal 5x. This can be solved through simple algebraic methods to give x=5 or x=0. 0 can be disagreed

Powers in GP, Infinite gp sum =1/1-1/2 =2 Root over 5^2 =5

let the given expration be x then we have x=sqrt(5x) or x^2 = 5x ==>x=0,5 hence the expresion is either zero or 5. Upvote iff agree..! :)

Expression = Product over n=1,2,3... of 5^(2^-n) = 5^(Sum over n=1,2,3... of 2^-n) = 5 ^ 1

y=Q Y y=5 Q y(y-5)=0 y=0,5

Assume that 5^1/2(5^1/2 5^1/2 ....= a ------(1) 5(5^1/2 5^1/2 .... = a^2 -------(2) I insert (1) in (2) will be 5a=a^2 So that a^2-5a=0 a(a-5)=0 a=5,0 but a cannot be 0 during question so a must be 5

Just made a brilliant thought. if there are five terms and they are all similar to the number of square root symbols, in this case 5, then the answer is 5.

y= sqrt(5* sqrt (5* sqrt (5* sqrt ... ... ... =>y^2= 5 y =>y^2-5 y =0 =>y*(y-5)=0 so, y=5

let root of 5...... be x

as the nested radical continues till infinity, it can be written as=

root of 5 *x = x

5x = x^2

x^2 - 5x =0,,, x(x-5) = 0,

x-5 = 0, x=5

x=√(5√(5√5) ) ……………∞, x^2=5x x=5

\sqrt { 5\sqrt { 5\sqrt { 5.. } } }

Lets assume: sqrt { 5\sqrt { 5.. } } = x 5x = { x }^{ 2 } { x }^{ 2 }-5x = 0 x(x-5) = 0 x = 0,5