You deal with them by checking that the values you found are actually solutions to the original equation. If you perform operations such as squaring both sides of an equation, there might be extraneous solutions, which won't be solutions of the original equation.How do we deal with missing solutions when we have to solve equations with non-reversible operations?
What do you mean by “things like” dividing by a variable that could be 0? The reason that I would have thought that dividing by a variable and using sin^-1 on both sides of an equation were “unsafe” is because neither are reversible operations. And I can’t think of a good example off the top of my head, but I feel like irreversible operations aren’t always avoidable.You deal with them by checking that the values you found are actually solutions to the original equation. If you perform operations such as squaring both sides of an equation, there might be extraneous solutions, which won't be solutions of the original equation.
You shouldn't really have "missing" solutions, unless you do things like dividing both sides by a variable whose value could possibly be zero. For example, if you have the equation ##x^2 = x##, it is tempting to divide both sides by x, getting the equation x = 1. That's not the smartest way to solve this equation, though. It's better to rewrite it as ##x^2 - x = 0##, and then factor the left side to ##x(x - 1) = 0##, from which you can obtain both solutions.
Another example is the equation ##\sin(x) = \frac 1 2##. If you naively apply the function ##\sin^{-1}## to both sides, you end up with ##x = \frac \pi 6##. Doing this, you miss out on ##x = \frac{11\pi} 6##, not to mention an infinite number of other solutions.
Or dividing by, say, x - 1 if x might be 1. There are lots of possibilities.What do you mean by “things like” dividing by a variable that could be 0?
You can always divide both sides of an equation by any nonzero quantity, but if you divide by a variable that could possibly be zero, then it's possible to lose solutions. The ##\sin## function is not 1-to-1, so it doesn't have an inverse that is itself a function. (We can, however, limit the domain such that ##\sin## is 1-to-1, but I wasn't doing that in the example I gave.)The reason that I would have thought that dividing by a variable and using sin^-1 on both sides of an equation were “unsafe” is because neither are reversible operations.
I can't think of any examples where you can't determine whether the operation is reversible. There are relatively few things that you can do to both sides of an equation: add/subtract the same quantity, multiply both sides by the same nonzero quantity, divide both sides by the same nonzero quantity, apply some function to both sides. If the function is 1-to-1 (i.e., has an inverse), then the step is reversible, and you won't have extraneous solutions or missing solutions.FAS1998 said:And I can’t think of a good example off the top of my head, but I feel like irreversible operations aren’t always avoidable.
I might add one to the list: substitution. Given any equation f=g involving well formed formulae f and g and given the equality x=y for variables (or well formed formulae that place no restrictions on the domain of their free variables) x and y, one can freely replace any occurrence of x in either f or g with y to obtain f'=g'.add/subtract the same quantity, multiply both sides by the same nonzero quantity, divide both sides by the same nonzero quantity, apply some function to both sides