SOLVE Hard Equation in real

Question

$$\sqrt { x + \sqrt { 2 x + \sqrt { 4 x + \sqrt { 8 x } } } } = \sqrt { 3 + \sqrt { 3 } }$$

First I squaring an equation but so many bushy of squaring And then got octic equation is hard to solve... Can anyone help me ???

Answer 1

By $f(x)$ we denote the LHS of the above equation ($x \ge 0$).

Then it is easy to see that $f(2) =\sqrt{3+\sqrt{3}}$. Since $f$ is strictly increasing on $[0, \infty)$, the equation $f(x)=\sqrt{3+\sqrt{3}}$ has the unique solution $x=2.$

Answer 2

The LHS is a monotonic function so that there is at most one solution.

You can suspect an integer solution and try $x=1$ then $x=2$ and bingo !

Alternatively, you can reason as follows:

As the RHS is made of two nested square roots with integers, and the LHS is a quadruple nesting, two unnestings must take place. One could be by the innermost argument $8x$ being a perfect square. Otherwise, two unnestings of the form $\sqrt{a+b\sqrt3}$ will be needed, and $x$ should be a multiple of $3$.

It doesn't take long to realize that $x=2$ has a likely order of magnitude, while $x=3$ or more would be too large. No other number than $2$ is possible.