Why do the number of solution to an equation reduces when we take square root on both the sides and increases when we exponentialize both the sides?

Question

My teacher says that when we are solving for $x^2=y$ then we have to put +- in front of y as solution to account for the loss of solutions that happened due to taking square root on both the sides.

Answer 1

Observe that both $x$ and $-x$, when squared, result in $x^2$.

For instance, the equation $x^2=4$ has both $x=2$ and $x=-2$ as solutions, because $$2^2={(-2)}^2=4$$

The $\pm$ is to account for this double solution that shows up naturally in the quadratic equation. For higher orders $(x^3,x^4,\dots)$ there are even more solutions.

Answer 2

If we have an equation $LHS = RHS$, then for each value of $x$, we get one value for $LHS$ and one value for $RHS$. We are looking for the $x$-s that make those two equal.

Now, if you square that equation to get $LHS^2 = RHS^2$, then the $x$-values that previously made $LHS$ and $RHS$ be the negative of eachother will now solve this new equation. Thus we (may) have created new solutions.

On the other hand, if we take the square root of the equation to get $\sqrt{LHS} = \sqrt{RHS}$, then the $x$-values that previously solved the equation, but happened to make the sides negative will no longer make sense. Thus we (may) have lost solutions.

These are the basics. Then there are the cases of when you square an equation with roots, or take square root in an equation with squares. In this case, we reverse the process above.

If we have $\sqrt{LHS} = \sqrt{RHS}$, then $x$-values that make $LHS = RHS$, but where they both are negative are not valid solutions. However, if you square that equation, then they become valid solutions.

If we have $LHS^2 = RHS^2$ then any any $x$-value that makes $LHS$ and $RHS$ equal is a solution, but also any $x$-value that makes one the negative of the other. If you take the square root on both sides, then such values of $x$ are no longer solutions. This is what is happening in your example.

There are some other nuances here, but I think this is enough for a general answer.