Kendall 501 Posted March 26, 2019 Report Share Posted March 26, 2019 All of the texts that I have looked at tell students to substitute the answer into the original to check for extraneous solutions. Aren't there some situations where you don't need to check, you just need to determine if the solution is in the domain? I'm thinking of functions with a variable in the denominator and logarithmic functions. In these two types of functions, is there a function for which this would not determine which ones were extraneous? Thanks in advance, Kendall Link to post Share on other sites

kiana 11,468 Posted March 26, 2019 Report Share Posted March 26, 2019 No -- the reason that radical equation could result in extraneous solutions which are still in the domain of the radical is because squaring (or raising to an even power in general) is not a one to one function. If you substitute an extraneous solution into a radical equation at the point at which you squared both sides, you will end up with something like -1 = 1, which of course is true when both sides are squared. The exponential function, however, is one to one, so it is impossible to have such things happen. If f(x) = a^x is an exponential function, and x is not equal to y, then there is no way for f(x) to equal f(y). The way an extraneous solution will happen is if you have something like log (x + 2) + log (x - 2) = log 5 and solve by combining to get log (x^2 - 4) = log 5. In this case, the extraneous solution x = -3 gives log (-1) + log (-5) = log 5, and you cannot use the logarithmic addition formula for inputs that are not in the domain. Multiplication is also one to one as long as whatever you are multiplying by is both defined and nonzero. With a rational equation that has an extraneous solution x = a, at some point you have multiplied both sides by (x - a), which is invalid when x = a -> x - a = 0. That being said, it's in the texts because it's a good habit to catch boneheaded mistakes 😛 Some students also find any kind of case-based "if a, do x, if b, do y" very confusing. 2 Link to post Share on other sites

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now