Are there any simpler ways to determine the solution for $\sqrt{x+\sqrt{x}}=1$ without back substitution checks?

Question

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?

Answer 1

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 $x=\frac{3-\sqrt5}{2}$.

Answer 2

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