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Teaching Precalc - extraneous roots?

  1. Sep 1, 2009 #1

    I am a first-year math grad student, and I just started teaching two sections of precalc! yikes!

    So I had a major embarrasing moment during office hours when I told 2 students the wrong answer because I completely forgot about the phenomenon known as "Extraneous roots"! Does anyone remember this from grade school?

    For example, if I have x^(1/4)=-2, putting both sides to the 4th power will give me x=16, but this is an incorrect solution, as 16^(1/4)=2. (EDIT: This is with the convention that roots are well-defined functions, so that the only give you the positive answer. I think if you see an equation to solve with radicals involved, it is assumed that the radicals mean positive roots). While this example is simple enough, there are cases where it is not obvious that I have an extraneous root until I plug-in to check!

    I am trying to create a list of operations that cause equations to gain additional roots (or lose them). So far, I have:

    Gain extraneous roots by:
    1. Multiplying both sides by an expression containing a variable. (common for solving rational equations).
    2. Raising both sides to an even power.

    Lose roots by:
    1. Dividing both sides by an expression containing a variable.
    2. Taking the 2n-th square root on both sides (unless you remember to put the obligatory plus-minus sign in front).

    Any others I am missing? I have consulted various algebra/precalc textbooks, but none of them give a satisfying theorem about this. Instead, they simply say "Make sure to check your answers, especially for solving equations involving rational polynomials or radicals".

    Also, is there a rigorous theorem/proof of when extraneous roots can appear? I feel like there should be a deep and general theorem about this in algebraic geometry.
    Last edited: Sep 1, 2009
  2. jcsd
  3. Sep 1, 2009 #2


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    A function f is injective if and only if every solution to f(x)=f(y) is also a solution to x=y.

    For example, the function f defined by
    f(t) = t²
    is not injective. Correspondingly, "squaring both sides" can potentailly introduce extraneous solutions.

    Similar statements can be made of f has extra parameters. e.g.

    Every solution to f(a,x)=f(a,y) is a solution to x=y if and only if, for each value of a, f(a,_) is an injective function.

    (f(a,_) is the function that maps x to f(a,x))

    The function f(s,t)=st violates this condition for s=0. Correspondingly, we see that multiplying both sides by an expression introduces extraneous roots iff 0 is in the image of that expression.
    Last edited: Sep 1, 2009
  4. Sep 1, 2009 #3
    Nice! It didn't occur to me to think in terms of functions. But how does "multiplying both sides by x" work? Because doing that sometimes introduces extraneous roots. (For example, multiply x=1 with x on both sides). But f(A)=xA is normally considered an injective function.

    And using the term "injective" is going to confuse students in precalc.
  5. Sep 1, 2009 #4


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    A partial function f is a total function if and only if every solution to x=y yields a solution to f(x)=f(y).

    For example, the square root function is not a total function. Thus, there are equations where taking the square root of both sides loses roots.

    Again, we can introduce parameters. f(s,t)=t/s is a partial function that isn't total -- it is not defined for any s=0. Therefore, we see that we lose roots wehenever we try to divide both sides by an expression whose image contains 0.
  6. Sep 1, 2009 #5


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    (Note I've added more to my first post)

    I'm not a teacher so my opinion might not count for much. But my opinion is that you should stress the operation they're actually doing.

    When they do something like
    x+6 = 2-x
    You should stress that what they have done is show that every solution to "x+6=2-x" is also a solution to "x=4" -- a lot of people understand this exactly backwards, and it's probably (at least partially) simply because nobody ever really explained it to them.

    As for the other problem, I opine it's because that it's because nobody ever drills into their heads that they really need to pay attention to the hypotheses. Many people look at:
    If z is nonzero, then from xz=yz we conclude x=y
    but what they remember is
    If xz=yz we conclude x=y

    How to actually teach this stuff? *shrug* Some pedantry might be worthwhile -- e.g. only give them partial credit when their work proves "If x is solution to x+6=2-x, then x=4" but didn't bother to prove "x=4 is a solution to x+6=2-x". (Or at least mention that each step is reversible, so that each step is actually an "if and only if")

    I'd be tempted to try and make little puzzles or games that demonstrate these two things. The first kind of puzzle would be solved by applying logic to narrow things down to a couple possibilities, and then actually checking each one to see what works. The second kind of puzzle is simply something that breaks down into multiple cases, and they have to work through each case individually.

    Solving the game of tic-tac-toe might be a fun and instructive exercise. :smile:
  7. Sep 1, 2009 #6
    The examples the OP mentioned are the ones that begining precalculus students should be aware of and will encounter. Extraneous roots also show up in logarithms when supposed solutions fall outside the domain of the equation.

    I have always found that discussing that the transformations made to an equation must preserve the solution set (i.e. reversable steps) and things like multiplying or divididing both sides of an equation by a variable expression, taking even powers or roots do not necessarily preserve the solution set are helpful.

    The mutplicative property says that provided c is nonzero then a = b implies ac = bc. If c is potentially zero then the transformation from a = b to ac = bc may introduce solutions (and going the other way may remove them). Squaring both sides is just a variation of multiplying both sides by a variable expression: a = b ==> ab = bb ==> aa = bb (by substitution).

    I encourage students to double check their solutions by verifying them in the original equation. In order to avoid losing solutions I stress that one should never divide by a variable quantity unless you are absolutely sure it is never zero (like e^x). Also, for even roots, I never let [itex]x^2= a[/tex] get short-cut to [itex]x= \pm \sqrt{(a)}[/itex] Students associate taking roots with the appearance of +/- and you get illogic like

    [tex]x^3=8 \Rightarrow x = \pm 2.[/tex]

    I hammer home the fact that

    [tex]\sqrt{x^2}=\left| x \right|[/tex]


    [tex]\left| x \right| = p \Rightarrow x = \pm p \text{ provided } p \geq 0 \text{ and unsolvable otherwise.}[/tex]

    (Side comment: turn your mistakes into a learning oportunity. Ask students to find the error and explain why it's an error - maybe even give them extra-credit for doing so. Keep them on their toes.)

    One way I have seen to make this process clear is the following:


    [tex]\Leftrightarrow \sqrt{x^3 - 4} + 10 - 10 = 12 -10[/tex]

    [tex]\Leftrightarrow\sqrt{x^3 - 4} = 2[/tex]

    [tex]\Rightarrow (\sqrt{x^3 - 4})^2 = 2^2[/tex]

    [tex]\Leftrightarrow x^3 - 4 = 4[/tex]

    [tex]\Leftrightarrow x^3 - 4 + 4 = 4 + 4[/tex]

    [tex]\Leftrightarrow x^3 = 8[/tex]

    [tex]\Leftrightarrow \sqrt[3]{x^3} = \sqrt[3]{8}[/tex]

    [tex]\Leftrightarrow x = 2[/tex]​

    Since not all rows are preceded by [itex]\Leftrightarrow[/itex] then the solution must be verified. In this example 2 is a solution, but if the original right-hand side were 8 then 2 would have been an extraneous solution.


    P.S.: The equation [itex]\ln{(x-1)} + \ln{(x+2)} = \ln 4[/itex] has an extraneous solution.
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