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Kendall

Extraneous solutions, do you really have to check?(math)

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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 

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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. 

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Yep, what @kiana said. The only time you don't need to check for extraneous solutions is if all of your manipulations can go backwards as well as forwards. So, for example, if you go from 

a = b + 1 to a - 1 = b,

that's a manipulation you can do either direction. But you can go from sqrt(a) = b to a = b^2, but you can't go backwards: b could be either sqrt(a) or -sqrt(a). And you can go from

x/a = b to ab = x, but not backwards,

because it's possible that a =  x = 0, and then you can't divide. 

It's kind of a pain to check whether every single step you do is reversible (although you get a feel for it if you do it enough). But generally, it's much easier to just plug back in. For another matter, plugging it back in connects what you're doing with what you're actually trying to find (a solution for the equation) and it's good for students to remind themselves what solving an equation really does :-). 

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