Where is the step that caused the extraneous solution?

Question

$$x^{1/3}+x^{1/6}-2=0\iff (x^{1/6}-1)(x^{1/6}+2)=0$$ $\iff x^{1/6}=1$ or $x^{1/6}=-2$

$\implies x=1^6=1$ or $x=(-2)^6=64$

By checking when $x=64$:$\sqrt[3]{64}+\sqrt[6]{64}-2=4+2-2=4\ne0$

Which step caused the extraneous solution $x=64$ to appear

Answer 1

An even root cannot have negative values so the solution $\sqrt[6]{x}=-2$ is not valid in $\mathbb{R}$

Answer 2

If you're restricting your attention to the real numbers, then how could you get $x^{1/6}<0$? It would be the same kind of thing if you tried to get a negative out of a square root: $x^{1/2}=\sqrt{x}<0.$ That is, there is no real number solution to $x^{1/6}+2=0$.

Answer 3

There is no real solution for $x$ in $x^{1/6}=-2.$ When you took the $6$th power of both sides, you lost information about the sign.

Answer 4

I'll take the algebraist's approach for this one.

The general rule is something like this: in an equation $L = R$ where $L$, $R$ are arbitrary terms, there are two general kinds of steps:

1. Rearraging one side

If you have an equation $$ L = R $$ then you can turn it into $$ L' = R $$ as long as $L$ and $L'$ are equivalent. You can do the same on the right-hand side, of course. This covers things like multiplying out brackets or taking a common denominator for fractions.

In your case, rewriting $x^{1/3} + x^{1/6} - 2$ as $(x^{1/6} + 2)(x^{1/6} - 1)$ is such a step.

This kind of step is always fine.

2. Applying a function $f$ to both sides.

Imagine you have an equation $x - 2 = 3$. Normally you'd "add 2 on both sides", what you're actually doing is applying the function $f(T) = T + 2$ on both sides (*1):

$$ \begin{array}{rrcl} & x - 2 & =& 3 \\ \Rightarrow & f(x-2) &=& f(3) \\ \Rightarrow & x - 2 + 2 &=& 3 + 2 \\ \Rightarrow & x &=& 5 \end{array} $$

The last step is a "rearranging" step on both sides.

The two most common examples of this case are $f(T) = T + C$ and $f(T) = TC$ where $C$ is any number (except $0$, in the second case), which corresponds to adding/subtracting and multiplying with a number to simplify the equation.

We have several cases for this technique:

• If $f$ is bijective (invertible), this kind of step is always fine. $f(T) = T + 2$ is bijective: its inverse is $f^{(-1)}(T) = T - 2$.

• If $f$ is not bijective, the step is still allowed but with caution, as it may cause extra solutions to appear, or lose some solutions (*2).

  The classic example of gaining extra solutions is applying $f(T) = T^2$ to $x = 2$ to get $x^2 = 4$ which has two solutions $2$ and $(-2)$. Conversely, applying $f(T) = \sqrt T$ to $x^2 = 4$ gets you $x = 2$, losing the negative root.

  In your case, applying $f(T) = T^6$ was the non-bijective function.

• (EDIT - adding this to avoid possible confusion) The function that you apply can itself contain the variable $x$. For example in the equation $x + 1 = 2x$ you want to "subtract an $x$ on both sides" with $f(T) = T - x$ to get $1 = x$. This function has the inverse $f^{(-1)}(T) = T + x$, whatever the value of $x$, so this step is fine.

  The place to be careful is when you have something like $$ \frac{1}{x} = \frac{x+1}{x} $$ The obvious function (*3) to choose is $f(T) = Tx$ getting you $1 = x + 1$. If $x$ is not $0$ then this function has an inverse $f^{(-1)}(T) = T/x$, whereas if $x$ turns out to be $0$ - as it does in this case - then you've multiplied by $0$, which is allowed but not very helpful as it gains you an extraneous solution.

  So if your function involves an $x$, you need to check whether it has an inverse for any possible $x$, and if not then check the cases when it doesn't again at the end.

  In this case, if you graph $1/x$ and $(x+1)/x$ you'll see that they don't intersect so there can't be any solutions to this equation; alternatively you could do a rearranging step to get $1/x = 1 + 1/x$ and then either conclude that nothing can ever equal itself plus one, or apply the function $f(T) = T - 1/x$ to get $0 = 1$ which also shows that there are no solutions.

Further notes

There are of course also non-general techniques that you need to solve certain kinds of equations, like splitting $(x-a)(x-b) = 0$ into two equations $x-a=0$ and $x-b=0$.

(*1) Experts in group theory are welcome to moan something about missing a step involving the associative law here.

(*2) To be precise, you can gain or lose solutions if the function is not bijective on the domain you're solving the equation for, in this case the reals. The problem with $\sqrt T$ is further that it's not even defined on all the reals, which is what loses you some solutions.

(*3) Alternatively you could directly choose $f(T) = Tx - 1$ to do "two steps at once". But that's not how they teach it in schools.

Answer 5

What's really going on here has nothing per se to do with the roots, contrary to what one might first suspect. If we take the elementary definition of rational powers (that a positive integral power of any real $x$ is a repeated product and a root of some nonnegative real $x$ is a number which multiplied repeatedly a certain number of times gives $x$), then in $\mathbb R$ $64$ is a perfectly valid (and the only solution of) $x^{1/6}=-2$ since $(-2)^6=64$ (this simply means $64$ is the only real number one of whose sixth roots is $-2$).

What we've been missing here is the fact that for any $a,b\in\mathbb R$, if $ab=0$, then $a=0$ or $b=0$. It is our deficient understanding of the nature of OR in mathematical logic that is causing all the confusion here. Or doesn't mean both $a$ and $b$ must vanish; on the contrary, it means either only one of them is $0$, OR (again) both of them are. So we're not always guaranteed that both of them are zero. It's sufficient that exactly one of them is. In usual cases when we solve quadratic equations by factorisation, both factors do indeed turn out to be zero, but that is not necessary in general. It's because we're used to this happenstance when the variables are simple and unadorned (i.e., they do not represent some mathematical expression different from the simple identity function of the variable) that we think every formal quadratic equation must behave likewise, but this is a logically baseless expectation.

In this case, the only solution of the original equation is $1$, as you can see easily by plotting its graph.

So the problem is simply logical interpretation. Some habits of normal life (wrong judgement) are hard to rub out, and are especially reinforced by schools, most of which teach mathematics mechanically.

PS. I think this teaches us the importance of checking whether possible solutions are indeed solutions, especially in cases where non-strict conditions like "if $ab=0$, then $a=0$ or $b=0$" are used, most especially when non-simple functions of the unknown are involved.