# Thread: Problem on square root...

1. ## Problem on square root...

Find the value of x in $\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=x$

I want to know where else I can find such questions and some hints to solve such questions if anyone can provide it would be a lot of help.

2. Hello siddscool19

I don't have an algebraic solution to this problem. If you try squaring both sides to get rid of square roots, you'll end up with an equation involving $x^{16}$. So I think you'll probably have to resort to a bit of intelligent trial and error.

If this equation is going to have any exact integer solutions then $3x$ will have to be a perfect square. The only candidates are $3x = 0, 9, 36, 81, ...$, giving possible values of $x$ as $0, 3, 12, 27, ...$

In fact $x = 0$ and $x = 3$ are solutions, and I'm certain they are the only ones.

I cheated a bit and used a spreadsheet to have a look at the function
$f(x)= \sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}$

When $0x$ and for $x > 3, f(x) and is increasing more slowly than $x$. So I'm sure that there are no further solutions.

I don't know whether anyone else can come up with a more analytical solution.

3. We have $x\geq 0$

A solution is $x=0$

Let $x>0$. Devide the equation by x:

$\displaystyle\sqrt{\frac{1}{x}+2\sqrt{\frac{1}{x^3 }+2\sqrt{\frac{1}{x^7}+2\sqrt{\frac{3}{x^{15}}}}}} =1$

Let $y=\frac{1}{x}$ and $f(y)=\sqrt{y+2\sqrt{y^3+2\sqrt{y^7+2\sqrt{3y^{15}} }}}-1$

f is strictly increasing and has a unique root: $y=\frac{1}{3}\Rightarrow x=3$

So, the solutions of the equation are $x=0, \ x=3$