Say we have: $$\sqrt{x+7}=5-x$$
Is it implicitly understood that the following also holds? $$-\sqrt{x+7}=5-x$$
I'm exploring the notion of "extraneous solutions." In this example, solving either equation leads to two results, namely x=2 and x=9.
Standard practice is to check these solutions once they're found by plugging them into the original equation. Thus, x=2 is shown to be the right answer, assuming we're using the first equation above. Meaning, x=9 is extraneous.
But really, x=9 is the solution to the second equation.
What's the right way to think about the square root operator given this discussion? Does the second equation logically follow from the first and, if so, is it right to call the second equation's solution "extraneous?"
Unless explicitely stated we only need to consider the positive square root on the left side. In fact the image of the square root function is the positive reals. Because of all this it's a common practice to plug in the solutions into the equation where we squared both sides.
For example in the formula for solving the quadratic equation we have $x_{1/2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, as only the positive value of the square root is taken.