# Solve the following equation: $\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} - \sqrt x= 1$

A past examination paper had the following question that I found interesting. I tried having a go at it but haven't come around with any solutions. How would one go about tackling it?

$$\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} - \sqrt x = 1$$

I'm seeing a relation between $4x$, $16x$ and $64x$ so maybe the larger can be simplified to the smaller?

I do encourage you to working within an examination environment (thus not making use of anything other than pen and paper and possibly a calculator).

EDIT: My question had a missing $-\sqrt x$ at the end, sorry!

• @JessePFrancis - Do take it in the case where one would be in the exam without the help of such widgets; what does the value imply? :-) – Juxhin May 22 '15 at 6:13
• I know, just commented it as a useful point, to verify if someone manages to come up with something! – Jesse P Francis May 22 '15 at 6:14
• @JessePFrancis - Oh sorry haha, wasn't aware. Thanks for the input! – Juxhin May 22 '15 at 6:16
• Did you write the question correctly? – mickep May 22 '15 at 6:24
• @mickep - I missed out a $-\sqrt x$ in all of that, has been added. Sorry :-) – Juxhin May 22 '15 at 6:27

$$\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} = 1+ \sqrt x$$

Squaring $$\sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} = 1+ 2\sqrt x$$

Squaring $$\sqrt {16x + \sqrt {64x + 5}} = 1+ 4\sqrt x$$

Squaring $$\sqrt {64x + 5} = 1+ 8\sqrt x$$

Squaring $$5 = 1+ 16\sqrt x$$

• There we go, makes alot more sense now! :-) – Juxhin May 22 '15 at 6:31
• @Juxhin, just missing a $\sqrt x$ gave enough training for the brain! – Jesse P Francis May 22 '15 at 6:33
• If anything I honestly appreciate how perfectly balanced this question is - was genuinely fun. Thanks again – Juxhin May 22 '15 at 6:34
• True that! When I saw the edit I was like, $\sqrt x$ just killed all the fun! The question does have a certain beauty worth admiring! – Jesse P Francis May 22 '15 at 6:38

This is a very intuitive approach based on the fact that I assume to be in the exam room with no computer and even no calculator.

If there is a simple root $\sqrt{64x+5}$ should reduce to a whole number and $x=\frac1{16}$ is obvious (since $9$ is the closest square to $5$). From here, we can go backward (verify that every time we get another square) and check that this is the solution.

• That's another simpler way to look at it; how did you deduce that $\sqrt {64x+5}$ was a simple root? – Juxhin May 22 '15 at 6:41
• I did not deduce it. I assumed that $\sqrt {64x+5}$ was a whole number and I went backward noticing that every time it was the case. – Claude Leibovici May 22 '15 at 6:43
• Oh alright that's a nice way to look at it – Juxhin May 22 '15 at 6:49
• Ah, the principle of well-stated questions comes in handy a lot, especially in competition and high-school maths. – MCT May 22 '15 at 6:50
• I guess that's what differentiates the great from the good.. – Juxhin May 22 '15 at 6:54

You get rid of the square roots by successive squarings and changes of side.

$$\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} = 1,$$ $$\sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} = -x+1,$$ $$\sqrt {16x + \sqrt {64x + 5}} = x^2-6x+1,$$ $$\sqrt {64x + 5} = x^4-12x^3+38x^2-28x+1,$$ $$0=x^8-24x^7+220x^6-968x^5+2118x^4-2152x^3+860x^2-120x-4.$$

Using a polynomial solver, there are six real roots and a complex conjugate pair, with no apparent simple value.

This makes the correctness of the problem statement rather dubious.

• Of course, one must check for extranneous solutions after solving. – MCT May 22 '15 at 6:20
• @Soke Right, for example $x$ can't obviously be greater than one. – Gregory Grant May 22 '15 at 6:26
• You're right Yves, the previous statement was incorrect. Has been updated, very sorry – Juxhin May 22 '15 at 6:32
• @Juxhin: no problem, nobody's perfect. Solving the bogus problem by hand seems out of reach. – Yves Daoust May 22 '15 at 6:35
• @YvesDaoust Touché – Juxhin May 22 '15 at 6:35

\begin{align*} \sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} - \sqrt x & = 1\\ \sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} & = 1+\sqrt{x}\\ x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} & = 1+x+2\sqrt{x}\\ \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} & = 1+2\sqrt{x}\\ 4x + \sqrt {16x + \sqrt {64x + 5}} & = 1+4x+4\sqrt{x}\\ \sqrt {16x + \sqrt {64x + 5}} & = 1+4\sqrt{x}\\ 16x + \sqrt {64x + 5} & = 1+16x+8\sqrt{x}\\ \sqrt {64x + 5} & = 1+8\sqrt{x}\\ 64x + 5 & = 1+64x+16\sqrt{x}\\ 16\sqrt{x} & = 4\\ x & =\frac{1}{16}. \end{align*}

The best solution I can imagine is to square repeatedly and then use numerical methods to find roots of the resulting polynomial and then checking for extranneous answers from our squaring.

• Same. The polynomial I got didn't factor nicely, though. I don't know how you'd solve it on an exam. – user231101 May 22 '15 at 6:17
• Hmm, I would end up getting; $x^4 + 4x^3 + 16x^2 + 64x + 5 = 1$ -- not sure if I did something incorrectly, I'll continue trying to factor it out. Do note that it's a 5 mark question in regards to a three hour exam which normally reflects how much time is expected to be allocated to the question – Juxhin May 22 '15 at 6:18
• @Juxhin You should get a degree $8$ polynomial, which has no general formula for solution. – MCT May 22 '15 at 6:19
• It's pretty cool though if you plug in $1/16$ how all those roots simplify to squares of rational numbers. – Gregory Grant May 22 '15 at 6:24
• @Soke, and then the little villain ($\sqrt x$) creeps in and kills all the fun! – Jesse P Francis May 22 '15 at 6:35

With the extra $\sqrt x$ there, you just square both side and keep doing it, miraculously some terms cancel out :)

• To be precise, move the square root term to RHS then square both side. – Chee Han May 22 '15 at 6:33