Equivalence of two radical equations without certain conditions - correctness of method

Question

I have question related to the following example: $\sqrt{22-x} - \sqrt{10-x}=2$.

First question:

Do I first need conditions $22-x \geq 0 $ and $ 10-x \geq 0$ to obtain the equivalent equation: $\sqrt{22-x}= \sqrt{10-x}+2$, then take square and so on or I don't need these conditions

i.e.$\sqrt{22-x} - \sqrt{10-x}=2$ is already equivalent to $\sqrt{22-x}= \sqrt{10-x}+2$, so we need only $10-x \geq 0$. The right side is non-negative, we take square and so on.

I wonder which method is correct.

Second question:

I know from theory that $\sqrt{f(x)}= g(x) \iff g(x) \geq 0 $ and $ f(x)= (g(x))^2 \geq 0$, so condition $f(x)\geq 0$ is automatically fulfilled. Can we do same here i.e. since right side is positive, so is left side in $\sqrt{22-x} - \sqrt{10-x}=2$ and we just take square of both sides?

Please help with this, I am learning for exams and I am confused about this. Thanks a lot in advance.

Answer 1

ANSWER TO 1ST QUESTION

I belive you can just mess around with the terms to find the solutions in $x$ but then you have to check if all of them are valid, ie you do need to use the conditions that $22-x\ge0$ and $10-x\ge0$. Is that helpful?

Answer 2

Your argument will be of the form $P_0(x) \implies P_1(x) \implies P_2(x) \implies \cdots$ and, hopefully, you will end up with a set of possible values for $x$, say $x \in D$. Essentially you can now say

$$\text{If $x$ solves $P_0(x)$, then $x \in D$.}$$

This is when you worry about what subset of $D$ will make $P_0(x)$ true.

There is an interesting "trick" that can be used to solve this equation.

\begin{align} \sqrt{22-x} - \sqrt{10-x} &= 2 \\ \sqrt{22-x} + \sqrt{10-x} &= y \\ \hline (\sqrt{22-x} - \sqrt{10-x})(\sqrt{22-x} + \sqrt{10-x}) &= 2y \\ (22-x)-(10-x) &= 2y \\ 12 &= 2y \\ y &= 6 \\ \hline (\sqrt{22-x} - \sqrt{10-x})+(\sqrt{22-x} + \sqrt{10-x}) &= 2+y \\ 2\sqrt{22-x} &= 8\\ \sqrt{22-x} &= 4 \\ 22-x &= 16 \\ x &= 6 \end{align}

And a quick check shows that $x=6$ is indeed a solution.