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I have solved the equation $$\sqrt{\frac{2}{x}}-\sqrt{\frac{x}{2}}=\frac{1}{\sqrt{2}}$$ where $x$ is $x=1$ and $x=4$. My question is why can't $x=4$? I understand that the equation does not hold if I put $4$ in the equation instead of $x$. Is there maybe a better and maybe a mathematical explanation for why $x$ can't be $4$?
Thank you in advance!
$x=4$ is not a solution to the equation because the equation is not true when $x=4$. There's nothing "better" or "more mathematical" than that as an answer to the question you asked. The question you should have asked is, "why did my solution process produce the incorrect solution $x=4$?". The answer to that might be that squaring is not a one-to-one function, so that squaring both sides of an equation (if that occurred in your solution process) can introduce spurious solutions.
Let's try to solve the following equation: $$x=1$$ Normal Way Obviously, the answer is just $x=1$.
Quadratic Way Square both sides: $$x^2=1$$ Subtract both sides by $1$: $$x^2-1=0$$ Factor and solve: $$(x+1)(x-1)=0 \implies x=-1 \text{ OR } x=1$$
In the normal way, we get the only correct solution. However, in the quadratic way, when we square both sides, we get an extraneous solution that is wrong. Often, when we square both sides this happens because, like in the example above, it'll count both positive and negative solution as the answer when really, we only need one of them. In the case above $x=1$ is the correct, positive solution where $\frac{1}{\sqrt{2}}$ is the answer on both sides while $x=4$ is the incorrect, negative solution where $-\frac{1}{\sqrt{2}}$ is on the left side, which is the negative of the right side.
Now, solve the following equation: $$\sqrt{\frac 2 x}-\sqrt{\frac x 2}=-\frac{1}{\sqrt 2}$$ Again, you should get $x=1$ and $x=4$. Which solution is correct this time? Why?
