Solution to $\sqrt{\sqrt{x + 5} + 5} = x$

Question

There are natural numbers $a$, $b$, and $c$ such that the solution to the equation \begin{equation*} \sqrt{\sqrt{x + 5} + 5} = x \end{equation*} is $\displaystyle{\frac{a + \sqrt{b}}{c}}$. Evaluate $a + b + c$.

I am not sure where I saw this problem. My guess is that it was from a high school math competition. The solution to the equation is $\frac{1 + \sqrt{21}}{2}$. This suggests use of the quadratic formula.

The solution set to the given equation is a subset of the solution set to \begin{equation*} x^{2} - 5 = \sqrt{x + 5} , \end{equation*} \begin{equation*} x^{4} - 10x^{2} + 25 = x + 5 \end{equation*} \begin{equation*} x^{4} - 10x^{2} - x + 20 = 0 . \end{equation*} Using the quartic equation (or Wolfram), the solutions to this equation are computed to be \begin{equation*} \frac{1 \pm \sqrt{21}}{2} , \qquad \frac{-1 \pm \sqrt{17}}{2} . \end{equation*}

Answer 1

Note that if $x = \sqrt{x+5}$ then $x = \sqrt{\sqrt{x+5}+5}$.

So, try solving $x = \sqrt{x+5}$. This is a quadratic.

Answer 2

• You can get the four apparent solutions from $x^{4} - 10x^{2} - x + 20 = (x^2 - x - 5) (x^2 + x - 4)$ and then solving two quadratic equations

• Squaring can produce spurious results:

  • for example if $x=\frac{-1 - \sqrt{17}}{2}$ then $\sqrt{\sqrt{x + 5} + 5} = -x$ rather than $+x$

  • but you lost that distinction when you went to ${\sqrt{x + 5} + 5} = x^2$

  • similarly $x=\frac{-1 + \sqrt{17}}{2}$ or $x=\frac{1 - \sqrt{21}}{2}$ then $\sqrt{x+5} = -(x^2-5)$ rather than $+(x^2-5)$ but you lost that distinction with the second squaring

When you use transformations which are not $1-1$ then you should check any apparent answers in the original question to see whether they are spurious or not

Answer 3

use the following way $$x=\sqrt{5+\sqrt{5 + x} }$$ $$x=\sqrt{5+\sqrt{5 + \sqrt{5+\sqrt{5 + \sqrt{5+\sqrt{5 + ....} }} }} } $$ or $$x=\sqrt{5+x }$$ $$x^2-x-5=0$$ $$x=\frac{1}{2}\pm\frac{\sqrt{21}}{2}$$ now use long division to get the other roots and then check which which one satisfies the original equation