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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!

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  • $\begingroup$ @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? :-) $\endgroup$ – Juxhin May 22 '15 at 6:13
  • $\begingroup$ I know, just commented it as a useful point, to verify if someone manages to come up with something! $\endgroup$ – Jesse P Francis May 22 '15 at 6:14
  • $\begingroup$ @JessePFrancis - Oh sorry haha, wasn't aware. Thanks for the input! $\endgroup$ – Juxhin May 22 '15 at 6:16
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    $\begingroup$ Did you write the question correctly? $\endgroup$ – mickep May 22 '15 at 6:24
  • $\begingroup$ @mickep - I missed out a $-\sqrt x$ in all of that, has been added. Sorry :-) $\endgroup$ – Juxhin May 22 '15 at 6:27
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$$\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} = 1+ \sqrt x$$

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

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  • $\begingroup$ There we go, makes alot more sense now! :-) $\endgroup$ – Juxhin May 22 '15 at 6:31
  • $\begingroup$ @Juxhin, just missing a $\sqrt x$ gave enough training for the brain! $\endgroup$ – Jesse P Francis May 22 '15 at 6:33
  • $\begingroup$ If anything I honestly appreciate how perfectly balanced this question is - was genuinely fun. Thanks again $\endgroup$ – Juxhin May 22 '15 at 6:34
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    $\begingroup$ 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! $\endgroup$ – Jesse P Francis May 22 '15 at 6:38
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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.

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  • $\begingroup$ That's another simpler way to look at it; how did you deduce that $\sqrt {64x+5}$ was a simple root? $\endgroup$ – Juxhin May 22 '15 at 6:41
  • $\begingroup$ 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. $\endgroup$ – Claude Leibovici May 22 '15 at 6:43
  • $\begingroup$ Oh alright that's a nice way to look at it $\endgroup$ – Juxhin May 22 '15 at 6:49
  • $\begingroup$ Ah, the principle of well-stated questions comes in handy a lot, especially in competition and high-school maths. $\endgroup$ – MCT May 22 '15 at 6:50
  • $\begingroup$ I guess that's what differentiates the great from the good.. $\endgroup$ – Juxhin May 22 '15 at 6:54
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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.

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

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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.

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  • $\begingroup$ Same. The polynomial I got didn't factor nicely, though. I don't know how you'd solve it on an exam. $\endgroup$ – user231101 May 22 '15 at 6:17
  • $\begingroup$ 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 $\endgroup$ – Juxhin May 22 '15 at 6:18
  • $\begingroup$ @Juxhin You should get a degree $8$ polynomial, which has no general formula for solution. $\endgroup$ – MCT May 22 '15 at 6:19
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    $\begingroup$ It's pretty cool though if you plug in $1/16$ how all those roots simplify to squares of rational numbers. $\endgroup$ – Gregory Grant May 22 '15 at 6:24
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    $\begingroup$ @Soke, and then the little villain ($\sqrt x$) creeps in and kills all the fun! $\endgroup$ – Jesse P Francis May 22 '15 at 6:35
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With the extra $\sqrt x$ there, you just square both side and keep doing it, miraculously some terms cancel out :)

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

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