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Solve for $x$: $$\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x$$

I tried to substitute $y=x+2$ and then I try to solve the equation by again and again squaring.

Then I got equation, $$(y-2)(3y^{14}-(y-2)^{15})=0$$

One solution is $y = 2$ and another is $y = 5.$ (I found $5$ as a solution of the equation by hit and trial method).

Therefore, $x = 0$ or $3.$

I'm wondering if there's any another method to solve it as the repeated squaring step seems to be somewhat absurd.

lone student
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  • If $x=\sqrt{x+2\sqrt{3x}}$, then we can replace $3x$ by $x+2\sqrt{....}$, and see it obeys the original equation. Untangling the square-roots gives $(x^2-x)^2=12x$, or $x(x-3)(x^2+x+4)=0$. So two more solutions are $-0.5\pm i\sqrt{15/4}$ – Empy2 Oct 14 '21 at 10:04
  • @Empy2 Your complex solution is not correct. – lone student Oct 14 '21 at 11:23
  • @lonestudent You have to pick the right square-roots. If $x=-\frac12+i\frac{\sqrt{15}}2$ then $\sqrt{3x}=-\frac32-i\frac{\sqrt{15}}2$, so $x+2\sqrt{3x}=-\frac72-i\frac{\sqrt{15}}2$, which is $x^2$. But then you have to pick the right square-root for the solution $x=3$. – Empy2 Oct 14 '21 at 12:12
  • @user979855 $y=5$ is not a solution of the equation $(y-2)(3y^{14}-(y-2)^{15})=0$ – perroquet Oct 20 '21 at 22:25

4 Answers4

2

It's not so hard to imagine someone guessing the solutions $0$ and $3$. The solution $0$ is something one can see from the positions of all of the "$x$". The solution $3$ might be inspired by asking what would make $\sqrt{3x}$ rational.

Now the idea is to prove that there can be no more than two solutions by showing that the left side is concave down. If $f(x)$ is positive and concave down, then first of all $2\sqrt{x+f(x)}$ is also positive. But:

$$\begin{align} \frac{d^2}{dx^2}2\sqrt{x+f(x)} &=\frac{d}{dx}\frac{1+f'(x)}{\sqrt{x+f(x)}}\\ &=\frac{f''(x)\sqrt{x+f(x)}-\frac{(1+f'(x))^2}{2\sqrt{x+f(x)}}}{(x+f(x))}\\ &=\frac{f''(x)(x+f(x))-\frac12(1+f'(x))^2}{(x+f(x))\sqrt{x+f(x)}} \end{align}$$

By assumption, $f''(x)$ is negative and $f(x)$ is positive, so this expression is also negative. So $2\sqrt{x+f(x)}$ is also concave down.

So since $2\sqrt{3x}$ is positive and concave down, so is $2\sqrt{x+2\sqrt{3x}}$. And so is $2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}$. And so is $2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}$.

Since that last expression $F(x)$ is concave down, there can be at most two solutions to $\frac12F(x)=x$.

2'5 9'2
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    Still I'm unable to understand how to solve for x in the given equation. –  Oct 14 '21 at 07:13
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    Couldn't one make this argument a bit quicker using the recursiveness of the equation? Each time we replace a $3x$ with a $x + 2\sqrt{3x}$ we can only create more solutions, never destroy them. However $(x + 2\sqrt{3x})/(3x)$ is a decreasing function for $x > 0$. So in that region we cannot create any new solutions, meaning that there can at most be one solution in the region $x > 0$. Moreover since $x \geq 0$ the only other possibility is $x = 0$. – Opisthokont Oct 14 '21 at 08:40
1

We have,

$$\begin{align}&\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x≥0\\ \iff &2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=2x\\ \iff &x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=3x\\ \iff &\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}=\sqrt{3x}\\ \iff &2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}=2\sqrt{3x}\\ \iff &x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}=x+2\sqrt{3x}\\ \iff &\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}} = \sqrt{x+2\sqrt{3x}}\end{align}$$

Let, $\sqrt{x+2\sqrt{3x}}=u,\thinspace u≥0$ then we get,

$$\begin{align}\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2u}}}}=u\end{align}$$

Then, we see that,

$$\begin{align}&\sqrt{x+2u} = u\\ \iff &\sqrt{x+2\sqrt{x+2u}} = u\\ \iff &\sqrt{x+2\sqrt{x+2\sqrt{x+2u}}}=u\\ \iff &\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2u}}}}= u\end{align}$$

Therefore, we have

$$\begin{align}&x+2u= x+2\sqrt{3x}\\ \iff &u^2=3x\\ \iff &x+2\sqrt{3x}=3x\\ \iff &\sqrt{3x}=x\\ \iff &x\in\left\{0,3\right\}.\end{align}$$

lone student
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1

We want to solve: $\ \sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=x $

So, we want to solve: $\ \sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2x}}}}=x \quad \quad (E)$

  1. We see that if $\sqrt{x+2x} =x$, then, $x$ is a solution of $(E)$. Therefore, $0$ and $3$ are two solutions of $(E)$.
  2. Suppose that $x$ is a solution of $(E)$ and $x\neq 0$. Then $x>0$ and $F(x)=1$ with $F(x)=\sqrt{\dfrac{1}{x}+\dfrac{2}{x}\sqrt{\dfrac{1}{x}+\dfrac{2}{x}\sqrt{ \dfrac{1}{x}+\dfrac{2}{x}\sqrt{\dfrac{1}{x}+\dfrac{2}{x}}}}}$
  3. $F$ is strictly decreasing on $(0,+\infty)$, so, the equation $F(x)=1$ has at most one solution.

We can conclude that $(E)$ has exactly two solutions, $0$ and $3$.

perroquet
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Set $F(x,y) = \sqrt{x+2y}$ for $x,y\geq 0$. Note that the function $F(x,y)$ is strictly increasing in $y$, i.e., if $y_1<y_2$, then $F(x,y_1)<F(x,y_2)$.

Set $$ y_0:=x; $$ $$ y_1:=F(x,y_0); $$ $$ y_2:=F(x,y_1); $$ $$ y_3:=F(x,y_2); $$ $$ y_4:=F(x,y_3), $$ and then, the equation is that $y_4 = y_0$.

Assume that $y_0<y_1$. Then $F(x,y_0) < F(x, y_1)$, i.e., $y_1<y_2$. Similarly, we get $y_2<y_3$ and $y_3<y_4$. I.e., in this case, we get that $y_0<y_1<y_2<y_3<y_4$, so the equality $y_4 = y_0$ cannot be satisfied.

Similarly, if we assume that $y_0>y_1$, then we get $y_0>y_1>y_2>y_3>y_4$, and again, the equality $y_4 = y_0$ cannot be satisfied.

Therefore, for the equation to be satisfied, $y_1$ has to be equal to $y_0$, i.e., $\sqrt{3x}=x$, which means that $x$ can be either $0$ or $3$. Substituting these values to the original equation, we see that both of them satisfy it.

Litho
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