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

Question

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).

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

Answer 1

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

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

Answer 2

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.

Answer 3

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.

Answer 4

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

Answer 5

\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*}

Answer 6

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.