While going over how to simplify expressions with radicals with my precalculus class, I ran into the problem of how to explain that $\sqrt[4]{9}$ can be simplified to $\sqrt{3}$. The best I could come up with was along the lines of you "just have to see that the fourth-root can be broken up into two square roots". So something like: $$ \sqrt[4]{9} = \sqrt[2]{\sqrt[2]{9}} = \sqrt[2]{3} \quad\quad\text{or}\quad\quad \sqrt[4]{9} = 9^{\frac{1}{4}} = (9^{\frac{1}{2}})^{\frac{1}{2}} = (3)^{\frac{1}{2}} = \sqrt{3} $$

The step of breaking up the one-fourth seems kinda tricky, and like something students wouldn't notice if they are in the mindset to simplify an expression. Is there a better way to explain this so that students will more immediately see that $\sqrt[4]{9}$ can be simplified? I am looking for an explanation that is more intuitive than just stating the rule, "If the power of any divisors of the radicand divides the index of the radical, then ...", because it seems like there should be some simpler explanation.

It might be better to say $$ \sqrt[4]{9} = 9^{\frac{1}{4}} = (3^{2})^{\frac{1}{4}} = (3)^{\frac{2}{4}} = (3)^{\frac{1}{2}} = \sqrt{3} $$ but turning the $9$ into $3^2$ will still run counter to the students' mindset of simplify.

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    $\begingroup$ is the square root of 3 really any more simple than the fourth root of 9? i don't mean to sound facetious but this is always something that has bothered me with high school curricula (including the district i work in), i.e having students "simplify" radicals when in the end they don't end up any simpler to work with except when in contrived, contextless problems, sorry... great question </rant> $\endgroup$
    – celeriko
    Oct 2, 2015 at 16:46
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    $\begingroup$ @celeriko there is some I think reasonable sense in which it is simpler, the former "is" the root of a degree 2 polynomial, while the latter "is" the root of a degree 4 polynomial. But generally I agree with you. $\endgroup$
    – quid
    Oct 2, 2015 at 17:04
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    $\begingroup$ @celeriko, I generally agree. I think the actual point of telling students to "simplify" an expression is to get them comfortable "playing around with" / manipulating an expression. That way when they (hopefully) get to calculus they are not bogged down with algebraic manipulations, but can instead focus on the concepts of calculus. It seems like it would be hard, though, to motivate a student to "play around" with an expression and notice all the equivalent ways to write it. The students need some goal to latch onto, so we introduce this ill-defined term "simplify". $\endgroup$ Oct 2, 2015 at 18:13
  • $\begingroup$ w.r.t. simplification: There are more remarks in this direction in MO 126519 (also mentioned in passing at the end of my response to MESE 2736). $\endgroup$ Oct 2, 2015 at 20:55

1 Answer 1


Which is simpler, $\sqrt[4]{243}$ or $3^{5/4}$ ? (Or even $3\sqrt[4]{3}$?)

The reason for teaching such rules at all is in the context of larger terms, where $\sqrt[4]{9}$ might cancel or combine with $\sqrt{3}$ in another place in the same or adjoining term. Further, it is useful to know some of the "notational patterns", because $9$ might be replaced by $(n^2 + 2n + 1)$ at some point.

Even if they are not going to be applying the rule often, there is a situation when they may need it. In this case noting that $\sqrt[stuff]{junk^{blah}}=junk^{blah/stuff}$ (with appropriate contextual restrictions) is really the relation to know, and your example of applying this relation to rewrite a term is one they may need to use.

Gerhard "Variable Names Need More Letters" Paseman, 2015.10.02

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    $\begingroup$ I do agree that trying to define "simpler" is silly. Unfortunately I have to abide by the curriculum which does try to define simplify. On a cute side note, I had quite a few students insist that $y^2\sqrt[3]{y^2}$ is simpler than $\sqrt[3]{y^8}$. $\endgroup$ Oct 2, 2015 at 18:21
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    $\begingroup$ @mapierce271, depending on context, it might well be... $\endgroup$
    – vonbrand
    Oct 5, 2015 at 22:23

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