I was trying to solve the equation, $x=1+\sqrt{x}$ for real $x$. Though I didn't correctly solve it. I'm curious as to why that is, and what else I need to initially consider in the domain of the function.

I started off by recognising that $x \geq 0$ for the square root to be real (I know when $x=0$ it is not a solution). Squaring both sides and rearranging; $$x^2 -3x + 1=0$$ Finding the solutions to this equation you obtain; $x=\frac{3\pm\sqrt{5}}{2}$. Both of these solutions to that equation are greater than zero, but only $x=\frac{3 + \sqrt{5}}{2}$ is the solution to the original. Why is that?

Is there some other "domain" restriction I must consider? Or for every question where there inolves root must I numerically test it (is there no way to get around this)?


  • $\begingroup$ Actually $x-1=\sqrt(x)≥0$ and hence x≥1 rather than 0. So there is only one root. $\endgroup$ – Huang Apr 29 '17 at 2:30
  • $\begingroup$ You x has to be one more than its square root. That's impossible if x is less than 1. $\endgroup$ – fleablood Apr 29 '17 at 2:30
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    $\begingroup$ The question is where does the extraneous solution slip in. Apparently not at x >=0. But first step is $x-1=\sqrt {x} >=0$ so.. $x >= 1$. So that's where it slips in. $\endgroup$ – fleablood Apr 29 '17 at 2:38

When you square an equation, you lose information. In your case, the equation $x -1 = \sqrt{x}$ will result in the same equation as $x-1 = -\sqrt{x}$ after squaring both sides: $(x-1)^2 = x$, regardless of the domain $x\geq 0$.

So your two roots $x = \frac{1}{2}(3 \pm \sqrt{5})$ come from one of them being a solution to $x-1 = -\sqrt{x}$ and the other $x-1 =\sqrt{x}$, even if both equations have domain $x \geq 0$.

For an answer to your last question, yes, in such cases, you'll need to check that any solution set you get from an equation after squaring it isn't a spurious solution by checking that it satisfies the original equation. To clarify this: it needn't be checking numerically - there are certain conditions you can check, for example as in law-of-fives comment, you have $x-1 =\sqrt{x} > 0$ so the only valid solution is the one that satisfies $x>1$.

  • $\begingroup$ It suffices to note that $\sqrt{x}$ must be positive given the original problem and when $x$ is found through the quadratic equation, given your first equation we not only have the criterion $x>0$ but also $x-1>0$. $\endgroup$ – law-of-fives Apr 29 '17 at 2:28

Let $y=\sqrt x\ge0$

$$\implies y^2=1+y\iff y^2-y-1=0$$

Clearly, the two roots are of opposite signs.

Now solve the quadratic equation.

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    $\begingroup$ This doesn't address the OP's question. $\endgroup$ – mrnovice Apr 29 '17 at 2:34

In general you "slip" the domain restiction at the exact point you square both sides.

$x=1+\sqrt {x} $

$x-1=\sqrt {x} $

!!! Here!! ===> $(x-1)^2=\sqrt {x}^2$<=== !!! Here!!! (so $x-1\ge 0$)

And with that in mind.....


The equation $x=1+\sqrt x$ can be rewritten as $x - \sqrt x - 1 = 0$. Hence the solution to the equation $x=1+\sqrt x$ is the root of the function

$$y = x - \sqrt x - 1 $$

Notice that $\sqrt x$ requires $x \ge 0$. The red plot below is the graph of the function $y = x - \sqrt x - 1$. We are interested in where this graph crosses the $x$-axis (since that's where $y=0$).

enter image description here

Let's rewrite the equation by eliminating the square root term.

\begin{align} y &= x - \sqrt x - 1 \\ \sqrt x &= x - y - 1 \\ x &= x^2 - 2xy + y^2 -2x + 2y + 1 \\ x^2 - 2xy + y^2 -3x + 2y + 1 &= 0 \end{align}

The equation

$$x^2 - 2xy + y^2 -3x + 2y + 1 = 0$$

is the equation of a parabola and is shown as a black dotted line on the plot above. Note that this equation is a extension (or a completion) of the original equation.

When you solved for $x$, you got the points $(\frac 12(3 \pm \sqrt 5), 0)$, one of which is on the red plot and one of which is not but both points are on the parabola.

Note also that

$$x^2 - 2xy + y^2 -3x + 2y + 1 = (x-y-1 - \sqrt x)(x-y-1 + \sqrt x)$$


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