# Find all real numbers $x$ such that $2+\sqrt{|x|}=x$

I'm revising for an exam and my lecture notes and online sheets etc. from my lecturers are very unhelpful, and I've completely forgotten how to do this type of question. So any help would be greatly appreciated!

The problem reads: "find all real numbers $x$ such that $2+\sqrt{|x|}=x$".

Thanks!

• After several edits it is still unclear for me what your one argument $\mathrm{mod}(x)$ should be?! Further: Do you mean $\left(\mathrm{mod}(x)\right)^{\frac{1}{2}}$ or $\mathrm{mod}(x^{\frac{1}{2}})$ or even (from your text) $\left(\mathrm{mod}(x)+2\right)^{\frac{1}{2}}?$ – gammatester Aug 13 '14 at 9:26
• What does mod mean to you? – Jyrki Lahtonen Aug 13 '14 at 9:26
• I mean 'the square root of the modulus of x'. I just don't know how to get the signs up properly. Have I got this mod confused with another? @gammatester – Charlie Robson Aug 13 '14 at 9:29
• So you mean $2 + \sqrt{|x|} = x?$ – gammatester Aug 13 '14 at 9:31
• Yes exactly that – Charlie Robson Aug 13 '14 at 9:32

The only solution to $2+\sqrt{|x|}=x$ is $x=4$.

Note that the graph of $y=\sqrt{|x|}+2$ is just the graph of $y=\sqrt{x}+2$ for $x\geq 0$ and $y=\sqrt{-x}+2$ for $x<0$.

$\sqrt{-x}+2=x$ has no solutions, so we have to look at $x=\sqrt{x}+2$ which has a solution at $x=4$.

$x=\sqrt{x}+2$

$x-2=\sqrt{x}$

$x^2-4x+4=x$

$x^2-5x+4=0$

$(x-4)(x-1)=0$

$x=4 \vee x=1$

$x=4$ is the only working solution.

• I was just going to upload that :-) so +1 – Freddy Aug 13 '14 at 9:42
• Graphs always come in handy! ;-) – rae306 Aug 13 '14 at 9:43
• This is a lot easier than I thought, thank you so much for your help! I Figured it would be more complicated than that!Thanks again – Charlie Robson Aug 13 '14 at 9:45
• Glad I could help :) – rae306 Aug 13 '14 at 9:45
• Alternatively, as we must have $x\ge0$, we can substitute $x=y^2$ with $y\ge0$ and get the quadratic $y^2=y+2$ with solutions $\frac12\pm\frac 32$, only one of which is $\ge0$ and leads to $x=2^2=4$. – Hagen von Eitzen Aug 13 '14 at 10:02

You have two cases - $|x|=x$ and $|x|=-x$ and you can also note that the square root is non-negative. This last observation means that $x\ge2$ so $x$ is positive.

This means that $|x|=x$ then you can rearrange and square to obtain $x=(x-2)^2$, which gives you a quadratic for $x$. You need to plug solutions back into the original equation to check them, because squaring can give rise to additional solutions and the solutions you find may not satisfy the conditions.

From $2 + \sqrt{|x|} = x\;$ you get $$\Longrightarrow\sqrt{|x|} = x-2$$ $$\Longrightarrow|x| = (x-2)^2$$ Now you must solve two restricted quadratic equations: $$(x-2)^2 = x,\quad x\ge 0$$ $$(x-2)^2 = -x,\quad x<0$$ The first has the two solutions $x_1=1, x_2=4\;$ and the second does not have real roots (only complex).

From the first pair only $x_2=4\;$ satisfies the original equation, and therefore the only solution is $x=4$.

You probably can't recall how to solve this as it doesn't have a classical form.

First identify the degree of the equation. The LHS includes a square root, i.e. degree $\frac12$; the RHS is of the first degree. This is not a polynomial equation, you need to reshape it for easier handling.

Get rid of the square root by isolating it and squaring: $$\sqrt{|x|}=x-2\\|x|=(x-2)^2.$$ Doing that, you possibly introduce an alien solution, to be rejected later by remembering that $x-2\ge0$.

Next, get rid of the absolute value. You do it by splitting in two cases: $$x<0\implies -x=(x-2)^2\\x\ge0\implies x=(x-2)^2.$$ Rearranging, you now have two nice quadratic equations for which standard formulas are available: $$x<0\implies x^2-3x+4=0\\x\ge0\implies x^2-5x+4=0.$$ After solving, check the constrains.

You could as well get rid of the absolute value by squaring, as $|x|^2=x^2$, but this would lead you to a scary quartic equation.

I really don't like the ${|x|}$, So what I do is I approach such problems like a child step-by-step as below.

\begin{align} 2 + \sqrt{|x|}= x\\\implies \sqrt{|x|}=x-2\\\implies{|x|}={(x-2)^2}\\ \end{align}

Case 1:

$+x=x^2-4x+4$

$\implies x^2-5x+4=0$

$\implies (x-4)(x-1)=0$

Case 2:

$-x=x^2-4x+4$

$\implies x^2-3x+4=0$

EDIT: After gammatesters highlighting the mistake, it appears that this will have imaginary roots.

So, that gives us $(4,1)$ as the possible solution.

However, When you do a Check, you will realize that it will pass only for $$x=4$$

• You have sign error $(-4)\times 1 = 4$ in case 2. There are only complex solutions. – gammatester Aug 13 '14 at 9:56
• You don't get x=-1 at all, it doesn't satisfy either quadratic, but I agree with the rest thank you – Charlie Robson Aug 13 '14 at 9:57
• No still wrong $(-4)\times 1\;$ remains $-4$ and not $4$ – gammatester Aug 13 '14 at 10:01
• @CharlieRobson With $\vert\,x\,\vert$ questions, you can get groups of solutions, not all of which are right. It's usually best to graph the problem first :) – Jam Aug 13 '14 at 10:02
• @gammatester Yes, sir! thank you for pointing it out. – MonK Aug 13 '14 at 10:04