Impossibly Long

Squaring both sides we get,

$x + \sqrt{x+\sqrt{x+\sqrt{\dots}}} = 25$

We know from the original equation that $\sqrt{x+\sqrt{x+\sqrt{x+\dots}}} = 5$ , so we substitute.

$x + 5 = 25 \Rightarrow x = \boxed{20}$

$\sqrt{x+\sqrt{x+\sqrt{x+....}}}=5$ .....(i)

On squaring both sides, we get,

$x+\sqrt{x+\sqrt{x+\sqrt{x+....}}}=25$

$x+5=25$ [From (i)]

$x=\boxed{20}$

We can consider a sequence as $a_{1} = \sqrt{x} , a_{2} = \sqrt{x+\sqrt{x}} , a_{3} = \sqrt{x+\sqrt{x+\sqrt{x}}} , ....$ Clearly for any $n>1 , a_{n+1} = \sqrt{x+a_{n}}$ that implies , $a^{2}_{n+1} = x + a_{n}$ We are given that $\lim_{n\rightarrow\infty} a_{n} = 5$ . So we take the both hand limits and get, $x = \lim_{n\rightarrow\infty} a^{2}_{n+1} - \lim_{n\rightarrow\infty} a_{n} = 5^{2} - 5 = 20$

Let $$(\sqrt ( x+\sqrt (x+ \sqrt (x+... \infty)=a ....(i) $$ Again $$(\sqrt ( x+\sqrt (x+ \sqrt (x+... \infty)=5 ....(ii) $$ Hence, $$a=5$$ Squaring on both sides of (ii): $$x+ (\sqrt ( x+\sqrt (x+ \sqrt (x+... \infty)=25$$ $$\Rightarrow x+a=25$$ $$\Rightarrow x=25-5$$ $$\Rightarrow x=20$$

We have: \begin{align} \sqrt{x + \sqrt{x + \sqrt{x + \dots}}} &= 5 \\ x + \sqrt{x + \sqrt{x + \dots}} &= 25 \\ x + 5 &= 25 \\ x &= \fbox{20} \end{align}

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

$\sqrt{x+5} = 5$

$x+5=25$

$x=\boxed{20}$

Given, $\sqrt{x + \sqrt{x + \sqrt{x + ...}}} = 5$ .......... (i)

Here, there are infinite $x$ terms inside the square roots.

Square both sides,

$x + \sqrt{x + \sqrt{x + \sqrt{x + ...}}} = 25$ .......... (ii)

Here, there are one less than infinite $x$ 's, which is same as infinite $x$ terms in the square roots.

Therefore, using (i), we can write (ii) as

$x + 5 = 25$

$x = 20$

That's the answer!

Squaring on both sides to the equation then we get $x+\sqrt{x+\sqrt{x+\dots}}=25$ Then substituting $\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}=5$ into the equation above, we will get $x+5=25$ . Finally we get $x=\boxed{20}$

√x+√x+√++........=√x+5=5(because it is an infinite series) squaring on both sides, x+5=25, x=20

Square both sides and subtract x from both sides:

$x+\sqrt{x+\sqrt{x+...}} = 25$

$\sqrt{x+\sqrt{x+...}} = 25 -x$

Then substitute in the original equation, and you have

$5 = 25-x$

$x=\boxed{20}$ .

$\sqrt{x+\sqrt{x+\sqrt{x+......}}}=5$ or $\sqrt{x+5}=5$ or $x+5=25$ or $x=\boxed{20}$

Let $\sqrt{x + .....}$ = a. Then, a = 5

Therefore, $\sqrt{x + a}$ = 5

Thus, x + a = 25

But a =5

Therefore, x = 20

The question is very easy,even though it looks tough $\large \sqrt{n+\sqrt{n+\sqrt{n+\dots}}} = 5$ And as the concept for such question is to use the formula $\large x = \frac{1+\sqrt{4n+1}}{2}$ Hence we do... $\large \frac{1+\sqrt{4n+1}}{2} = 5$ We multiply RHS and LHS by 2 $\large 1+\sqrt{4n+1} = 10$ Then we subtract 1 from RHS and LHS $\large \sqrt{4n+1} = 9$ Squaring both the sides $\large 4n+1 = 81$ Subtracting 1 from both sides $\large 4n = 80$ Dividing both sides with 4 $\large n = 20$

That 8 legged cat made my little brother laugh out of his wits....! Awesome question and the apt (octapede... :-) )image'-.....!

I had no idea what I was doing so I first converted it to $X^{1/2}+X^{1/4}+X^{1/8}...=5$

Then I realized that $X+X^{1/2}+X^{1/4}+X^{1/8}...=5^{2}$

and that $X+X^{1/2}+X^{1/4}+X^{1/8}...=5+X$

So I just did $25=X+5$

Let y = √ x + √ x + √ x + √ x +……………….. Now according the question y = 5 y^2 = 25 = x + √ x + √ x + √ x +……………….. or we can say y^2 = y + x 25 = 5 + x or x = 20

root inside root go to infinite times then we assume , if we hold a first root x and until go to 5 because this turns TO INFINITE. let's simplfies sqrt(x+5)=5; x+5=25; x=20;

\sqrt{x}+\sqrt{x}+\sqrt{x}..........=5 \sqrt{x}+5 =5 Squaring bothsides, x+5=23 Therefore,x=20

In the equation, keep the roots other than the value of the first 'x' as 5. Solving, root of (x+5)=5 Squaring, we get, x+5=25.So,x=20.

√(x +√(x + √(x + . . . . .))) = 5 ---------(i) Squaring both sides, we get x +√(x + √(x + . . . . .)) = 25 or, x + 5 = 25 [from (i)] So, x = 25 – 5 = 20

square on both sides you will get

x + sqrt(x(sqrt(xsqrt(x)))..= 25 , Now the value of sqrt(x(sqrt(xsqrt(x))) = 5 from given question. So replace it with 5.

you will get x + 5 = 25 , which implies x = 20

Solution: √(x+√(x+√(x+⋯)))=5 ≫√(x+5)=5 [ as √(x+√(x+√(x+⋯)))=5 ] ≫x+5=25 ∴x=20

Square the expression on both the sides. One x will come out. Replace the remaining x's with 5 as they continue till infinite. You will have $x+5=25$ . Hence $x=20$ .

x=20

√(x+√(x+√(x+⋯)) ) =5 〖(√(x+√(x+√(x+⋯)) ) )〗^2=5^2 x+√(x+√(x+⋯)) =25 As we know that √(x+√(x+√(x+⋯)) ) =5 x+5=25 x=25-5 x=20

20

y = sqrt(x + sqrt(x + sqrt(x + ….)))

y2 = x + sqrt(x + sqrt(x + sqrt(x + ….)))

y2 = x + y

Now, we have y = 5

52 = x + 5

25 = x + 5

x = 20

GIVEN THAT,

√(x+√(x+√(x+⋯)=5

SO WE CAN WRITE THAT,

√(x+5)=5

NOW,

x+5=5^2

x+5=25

THEREFORE, x=25-5=20

squaring both sides, then put \sqrt{x}+\sqrt{x}+....+5 then it will give us x=20 Ans

Since infinite number of square roots are there, we approach the prob like this : 1. Square on both sides 2 . x + infinitely long term = 25 3 . Since, infinitely long term = 5 4 . x + 5 = 25 5 . x = 20

The above problem can be written as sqrt{x+5} = 5; or, x+5 = 25; x= 25-5; or x = 20

Squaring both sides, we get x+ \sqrt{x+\sqrt{x+...}} = 25. Using the given equation, x+5=25. Hence, x=20.

at first consider the whole expression to be y=5 next ,square both sides..we get. x+y=25 since y=5 so x=20

\sqrt x+5=5

squaring both side

x+5=25

x=20

given root of(x+<x+.....)=5 implies root of(x+5)=5 implies x+5=25 then we get x=20

square the both side of the given equation and then we get that x+5 = 5^2 =25 implies that x=20

Based on my analysis, on every radicand must be equal to 25 to become 5, if you subtract 5 from 25 it become 20, so that your x is 20.

squaring both sides: x+5=25; x=20

let the left side of the equation be =a squaring on both sides we get x+a=25 as a=5 x=20

√(x+√(x+√(x+⋯)) ) = 5

[√x+√(x+√(x+⋯)) ]^2= [5]^2

Let 5 be them:

x+(5)=25 x=25-5 x=20

x + 5 = 25, x = 20.

squaring both sides we get: x +y=25 (y=l.h.s.=5 which is the initial eqn.) x+5=25 x=20

Here, $\sqrt{x+\sqrt{x+.....}+.....}$ =5

Hence We have,

$\sqrt{x+5}$ =5

So, x+5=25

Or, x=20

The limit of $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ is 5

Hence, we have

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

Solving for x gives $\boxed{x = 20}$

25-5=20

sqrt(x + sqrt(...))=5 x+sqrt(...)=25 5=sqrt(...) x+sqrt(...)-sqrt(...)=25-5=20

it is infinitely long , so if we take out one x , it would not affect the whole series as the series is infinitely long. so the remaining series becomes exactly the same series which is equal to 5.

so therefore

5= Vx+5 squaring and solving , we get x=20

$\sqrt(x+\sqrt(x+\sqrt(x+...)))=5\Leftrightarrow\sqrt(x+5)=5\Leftrightarrow x+5=25\Leftrightarrow x=20$

Notice that $\sqrt{x+5}=5$ . Squaring both sides, we get $x+5=25$ so $x=\fbox{20}$ .

tke infinite =5 .. what we left is x+5=5^2 ; 20

5=sqrt(x+5) => 25=x+5=> 20=x.

We define the expression with roots to be S -> S=5. By squaring both sides we get x+S=25, and thus x=25-5=20

square both sides... Now the eqn becomes x+5=25 ( putting given eqn again) So, x=20

square both sides 25=x+5

Let $\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$ denote the infinite nested radicals.

We have $\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}} = 5 \Rightarrow x + \sqrt{x+\sqrt{x+\sqrt{x+\cdots}}} = 5^2$ . But because $\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}} = 5$ , we can substitue it back and find that $x + 5 = 25 \Rightarrow \boxed{x = 20.}$