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A weak condition by inspection: $x>0$.

\begin{gather} \sqrt{x+\sqrt x} = 1\\ x+\sqrt x = 1\\ \sqrt x = 1-x\\ x = 1-2x+x^2\\ x^2 - 3x + 1 =0\\ x=\frac{3\pm\sqrt5}{2} \end{gather}

As both satisfy the weak condition $x>0$ it seems to me both are the solution. However by tedious back substitution only $x=\frac{3-\sqrt5}{2}$ is the correct solution.

Question

Are there any simple approaches to find the solution but without back substitution?

D G
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    All you need to consider is the line you squared, $\sqrt x = 1-x$. The apparent root $\frac {3+\sqrt 5}2$ clearly doesn't satisfy this, so it can be discarded. It's the squaring that introduces false solutions, as you risk losing a sign. – lulu Apr 25 '23 at 18:50
  • There is a another easy way. We have $x=(3+\sqrt{5})/2\sim 2.618$, and for this $x$ then $\sqrt{x+\sqrt x}\sim 2.058$, which is far away from $1$. – Dietrich Burde Apr 25 '23 at 18:51
  • Or just consider $x>\frac 32>1$ and $\sqrt{x+\sqrt{x}}>\sqrt{2}>1$ – zwim Apr 25 '23 at 18:51
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    Whenever you square an equation where one side is positive, add the condition that the other side must be positive, too. This means that before you square for the second time, you write down the condition $1-x \ge 0 \iff x \le 1$, then at the end you use it eliminate the extraneous root. In other words, $\sqrt{u}=v \iff v \ge 0 ,\land, u = v^2$. – dxiv Apr 25 '23 at 18:53
  • @dxiv Your comment is the one at the root of the question! – mathcounterexamples.net Apr 25 '23 at 18:57

2 Answers2

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The implication $a = b \implies a^2 = b^2$ is valid in general. But the reverse implication is valid only if the additional condition $\operatorname {sgn}(a) =\operatorname {sgn}(b)$ is true.

Your inequalities are implicited chained by $\implies$. To turn then into $\iff$ you need:

  • $1 \iff 2$ : true if $x + \sqrt{x} >0$ (true, given that $x>0$)
  • $2 \iff 3$ : true
  • $3 \iff 4$ : only true if $1-x >0$; that is, $x<1$
  • $4 \iff 5$ : true
  • $5 \iff 6$ : true

Then to go back to the first one you can write

$$ x=\frac{3\pm\sqrt5}{2} \cap x<1 \implies \sqrt{x+\sqrt x} = 1$$

which reduces to the single root $x=\frac{3-\sqrt5}{2}$.

leonbloy
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A bit of speedcheating the problem...

You can use the fact that $x$ and $\sqrt{x}$ are continuous and strictly increasing functions on $\mathbb R^+$ therefore the equation is bijective and there can only be one unique solution.

Now equations of such form often involve the golden ratio.

Remember indeed that $\phi^2=\phi+1$ or equivalently $1=\underbrace{\frac 1\phi}_{\sqrt{x}}+\underbrace{\frac 1{\phi^2}}_x$

And there you have it $x=\dfrac 1{\phi^2}=(\bar\phi)^2=\left(\dfrac{1-\sqrt{5}}2\right)^2=\dfrac{3-\sqrt{5}}2$

zwim
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