Is $x=-2$ a solution of the equation $\sqrt{2-x}=x$?

Question

Solve the equation:

$$\sqrt{2-x}=x$$

Squaring we get

$$x^2+x-2=0$$ So

$x=1$ and $x=-2$

But when $x=-2$ we get

$$\sqrt{4}=-2$$

But according to algebra $$\sqrt{x^2}=|x|$$

So is $x=-2$ invalid?

Answer 1

Yes, $x=-2$ is invalid because $$x=\sqrt{2-x}\geq0.$$

I like the following reasoning.

$x$ increases, $\sqrt{2-x}$ decreases, which says that our equation has one root maximum.

But $1$ is a root and we are done!

Answer 2

Keep in mind, that 'squaring' is not an equivalent term transformation. It manipulates the domain of your equation and can therefor create 'fake solutions'.

So either you have to be more careful, when squaring, or you have to check every solution if it is fake or not. (Which means that the original equation holds true for this input)

In this case $x=-2$ is a fake solution and $x=1$ is a real solution to your equation.

Answer 3

When you square the equation, you add a solution you didn't have before. Take for example the very simple equation - $$x=1$$ Squaring both sides we get: $$x^2=1\Rightarrow x=\pm 1$$ Not all operations leave the information given in the original equation unchanged. Just like if you multiply an equation by $0$ you get an equation that's always true ($0=0$), but you basically lose all information, when squaring an equation, you can sometimes add a new solution.

Answer 4

Before doing anything, you know that $x\ge0$ because the real square root function is non-negative.

By inspection $x=1$, which is the only solution.

Answer 5

$$ \sqrt{2-x}=x. $$

There are two things that we can glean from that statement above: $2-x\ge0$ and $x\ge0$. The expression under the square root must be greater or equal to zero and if something equals the square root of something else, it must also be greater than or equal to zero. All this means that the statement above only makes sense if $0\le x\le 2$. Something that does not belong to that interval is a so-called extraneous solution and must be discarded. It might be a solution to an equation you get as an intermediary step, but not to the original equation. More properly, your solution process should really be looking something like this:

$$ \sqrt{2-x}=x\implies\\ x^2+x-2=0,\ 0\le x\le 2\implies\\ x=1. $$

Even though the equation $x^2+x-2=0$ has two roots, $-2$ and $1$, the equation $x^2+x-2=0,\ 0\le x\le 2$, which is equivalent to your original equation $\sqrt{2-x}=x$, has only one root, $1$. That's simply because there is a restriction on what values of $x$ can be part of the solution set.

Answer 6

In your title you ask, is $x = -2$ a solution of $\sqrt{2-x} = x$?

To answer this question, you simple need to check the answer by substituting $x=-2$ into the original question and see if a true statement results. In this case, we get

$$ \sqrt{2-(-2)} = \sqrt{4} = 2$$

which does not equal the RHS $x = -2$.

Therefore $x=-2$ is not a solution of the original equation.

REMARK: Someone may try to argue that $-2$ is a square root of $4$ because $(-2)^{2}$ is $4$; and indeed it is. But when a term of an equation is written as $\sqrt{2-x}$, it means to take the positive square root of $2-x$; otherwise, if the term would be written as $-\sqrt{2-x}$ if the negative square root was meant. (Just like $\sqrt{4} = 2$ and $-\sqrt{4} = -2$.)