Find the value of x in $\displaystyle \sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=x$
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.
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 $\displaystyle x^{16}$. 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 $\displaystyle 3x$ will have to be a perfect square. The only candidates are $\displaystyle 3x = 0, 9, 36, 81, ...$, giving possible values of $\displaystyle x$ as $\displaystyle 0, 3, 12, 27, ...$
In fact $\displaystyle x = 0$ and $\displaystyle x = 3$ 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
$\displaystyle f(x)= \sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}$
When $\displaystyle 0<x<3, f(x)>x$ and for $\displaystyle x > 3, f(x)<x$ and is increasing more slowly than $\displaystyle x$. 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.
Grandad
We have $\displaystyle x\geq 0$
A solution is $\displaystyle x=0$
Let $\displaystyle x>0$. Devide the equation by x:
$\displaystyle \displaystyle\sqrt{\frac{1}{x}+2\sqrt{\frac{1}{x^3 }+2\sqrt{\frac{1}{x^7}+2\sqrt{\frac{3}{x^{15}}}}}} =1$
Let $\displaystyle y=\frac{1}{x}$ and $\displaystyle f(y)=\sqrt{y+2\sqrt{y^3+2\sqrt{y^7+2\sqrt{3y^{15}} }}}-1$
f is strictly increasing and has a unique root: $\displaystyle y=\frac{1}{3}\Rightarrow x=3$
So, the solutions of the equation are $\displaystyle x=0, \ x=3$