Suppose $x$ is a real number such that $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x.$ Find $x.$

I first squared to get rid of the first square root, which gave me $\sqrt{3} - \sqrt{\sqrt{3} + x} = x^2.$ However, I'm not sure how to move on from there. Can someone give me a hint?

  • 4
    $\begingroup$ Hint: subtract $\sqrt{3}$ from both sides and square $\endgroup$
    – QC_QAOA
    Sep 17 '20 at 15:35

From where you left off:

$$\begin{align} \sqrt{3}-\sqrt{\sqrt3+x} &=x^2\\ \sqrt{3}-x^2 &=\sqrt{\sqrt3+x}\\ 3-2\sqrt{3}x^2+x^4 &=\sqrt{3}+x\\ x^4-2\sqrt{3}x^2-x+\left(3-\sqrt3\right)&=0\\ \left(x^2+x+(1-\sqrt{3})\right)\left(x^2-x-\sqrt{3}\right)&=0 \end{align}$$

So if there is a solution to the original equation, it is a root of this 4th degree polynomial. It's easy to find its four roots since it factors. But some roots of this polynomial might not solve the original equation, since we squared a few times earlier. So each one should be checked.

Note that the original left-hand side is not real unless $x$ is in $\left[-\sqrt{3},3-\sqrt{3}\right]$. That should help eliminate several of the four polynomial roots.


Let $a = \sqrt{3}$, so $x = \sqrt{a-\sqrt{a+x}}$. Let's call this equation $(*)$.

Let $y = \sqrt{a+x}$ so $x = \sqrt{a-y}$. Therefore

  1. $x^2 = a-y$
  2. $y^2 = a+x$

Subtracting 2. from 1. we have $x^2 - y^2 = -(x+y)$. Let's consider the following cases:

  • If $x+y = 0$, then $x = -\sqrt{a+x}\le 0$, but from $(*)$ we have $x\ge 0$, so we must have $x = 0$. It's easy to see that this doesn't work.

  • If $x+y \neq 0$, we have $\begin{aligned}\frac{x^2-y^2}{x+y} = -1 \therefore x-y = -1\end{aligned}$, i.e., $x+1 = \sqrt{a+x}$, so $(x+1)^2 = x^2+2x+1 = a+x$, therefore $x^2+x+1-a = 0$, which can be solved as a quadratic equation to get $\begin{aligned}x = \frac{-1+ \sqrt{1-4(1-a)}}{2} = \frac{-1+\sqrt{4a-3}}{2}\end{aligned}$ (notice that this is the only positive solution).

  • $\begingroup$ y=sqrt(a+x) doesnt imply x=sqrt(a-y) $\endgroup$ Sep 17 '20 at 15:56
  • 2
    $\begingroup$ @SoumyadwipChanda, it does. Just replace $y$ in the original equation. $\endgroup$
    – Arjuna196
    Sep 17 '20 at 16:00


If you put $x=\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} $ under the third radical you get:

$$\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + \sqrt{\sqrt{3} - \sqrt{\sqrt{3} + \sqrt{\sqrt{3} - \sqrt{\sqrt{3} + }} . . .}} }} = x$$

Suppose there is no negative sign then by squaring both sides you get:

$$X^2=\sqrt 3 +X$$

Which gives:

$$X=\frac{-1 ±\sqrt{1+4\sqrt 3}}{2}≈0.9≈\frac{\sqrt3}{2}$$

So $x<X≈\frac{\sqrt 3}2$

  • $\begingroup$ "Squaring both sides you get:" how, exactly, do you square the right side which is an infinite expression? Simply squaring it would just remove the outermost root. $\endgroup$ Sep 17 '20 at 16:07
  • $\begingroup$ When you square the right side, you do not get $\sqrt{3}-x$ because the signs ($+$ and $-$) are shifted. $\endgroup$ Sep 17 '20 at 16:08

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