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Solve for $x$:


My attempt:

The L.H.S is equal to $\dfrac{1+\sqrt{4x+1}}{2}$ and R.H.S equals $x^2$

Equating both sides:

$\implies 4x+1=(2x^2-1)^2$

$\implies 4x+1=4x^4-4x^2+1$

$\implies 4x^4-4x^2-4x=0$

$\implies x(x^3-x-1)=0$

Disregarding the complex roots,

$\implies x=0$ 0r $\dfrac{1}{3}\sqrt[3]{\dfrac{27-3\sqrt{69}}{2}}+\dfrac{\sqrt[3]{\dfrac{9+\sqrt{69}}{2}}}{3^{2/3}}$

Is my solution correct? By the way I would like to see other methods to solve it. Thanks!

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I am not getting that the RHS = x^2 – bobbym Aug 5 '14 at 13:40
$\sqrt{xa}=a \implies \sqrt{x} = \sqrt{a} \implies x=a$, i.e. R.H.S = $x$. – Daniel R Aug 5 '14 at 13:42
Those exponents sum to 1 not 2. I think the RHS = x. – bobbym Aug 5 '14 at 13:42
We have 62 questions tagged nested-radicals, many of which are relevant to your question. – MJD Aug 5 '14 at 13:44
Not that it matters, but the complex form can be rewritten as: $$\frac{1}{3}\left(\sqrt[3]{\dfrac{27-3\sqrt{69}}{2}} + \sqrt[3]{\dfrac{27+3\sqrt{69}}{2}}\right)$$ which is a little clearer. It doesn't matter because of the error evaluating the RHS. – Thomas Andrews Aug 5 '14 at 13:47
up vote 11 down vote accepted

I get a different solution:

Let $y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+..}}}}=\sqrt{x\sqrt{x\sqrt{x\sqrt{x...}}}}$.

Then $y^2 = x+ y$ and $y^2 = xy$.

From the second equation we see that $x = y = 0$ is a possible solution, otherwise, $x = y \neq 0$.

So from the first equation: $x^2 = 2x \Rightarrow x = y = 2$

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I'd also got that. The way the asker's done it, it should be $x\sqrt{x\sqrt{x\sqrt{...}}}$ on the RHS to get their solution. – Eul Can Aug 5 '14 at 13:43
My God!! What have I done!!!!!!! – Tom Lynd Aug 5 '14 at 13:44
From the second equation we got $x = y$ so I sustituted $y$ for $x$ on the first equation. – Darth Geek Aug 5 '14 at 13:47

Note that RHS is wrong: $$y=\sqrt{x\sqrt {x\cdots}}\implies y^2=x\sqrt{x\sqrt {x\cdots}}=xy\implies y=0 \text{ or }x$$

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