(I've been posting a lot today and yesterday, not sure if too many posts are frowned upon or not. I am studying and making sincere efforts to solve on my own and only post here as a last resort)

I'm asked to simplify $\sqrt[4]{\frac{162x^6}{16x^4}}$ and am provided the text book solution $\frac{3\sqrt[4]{2x^2}}{2}$.

I arrived at $\frac{3\sqrt[4]{2x^6}}{2x^4}$. I cannot tell if this is right and that the provided solution is just a further simplification of where I've gotten to, or if I'm off track entirely.

Here is my working:

$\sqrt[4]{\frac{162x^6}{16x^4}}$ = $\frac{\sqrt[4]{162x^6}}{\sqrt[4]{16x^4}}$

Denominator: $\sqrt[4]{16x^4}$ I think can be simplified to $2x^4$ since $2^4$ = 16

Numerator: $\sqrt[4]{162x^6}$ I was able to simplify (or over complicate) to $3\sqrt[4]{2}\sqrt[4]{x^6}$ since:

$\sqrt[4]{162x^6}$ = $\sqrt[4]{81}$ * $\sqrt[4]{2}$ * $\sqrt[4]{x^6}$ = $3 * \sqrt[4]{2} * \sqrt[4]{x^6}$

Thus I got: $\frac{3\sqrt[4]{2}\sqrt[4]{x^6}}{2x^4}$ which I think is equal to $\frac{3\sqrt[4]{2x^6}}{2x^4}$ (product of the radicals in the numerator).

How ca I arrive at the provided solution $\frac{3\sqrt[4]{2x^2}}{2}$?

  • $\begingroup$ Hint, $\sqrt[4]{16x^4}$ does not simplify to $2x^4$. Rather, it becomes $2x$. $\endgroup$ Jan 6, 2019 at 21:05
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    $\begingroup$ On your first comment. Posting a lot is fine as long as you are actually working on each of the questions you ask (as you clearly are on this one), $\endgroup$ Jan 6, 2019 at 21:13

2 Answers 2




You made an error:

$$\sqrt[4]{16x^4} = 2\vert x\vert$$

because $\sqrt[4]{16x^4} = \sqrt[4]{(2x)^4}$. (Note the absolute value sign since the value returned is positive regardless of whether $x$ itself is positive or negative.) The rest is fine, so from here, you get

$$\frac{3\sqrt[4]{2x^6}}{2\vert x\vert} = \frac{3\sqrt[4]{2x^4x^2}}{2x} = \frac{3\vert x\vert\sqrt[4]{2x^2}}{2\vert x\vert} = \frac{3\sqrt[4]{2x^2}}{2}$$

As shown in the other answer, it is usually better to simplify within the radical so you don’t mess up with absolute values (for even indices).

$$\sqrt[4]{\frac{162x^6}{16x^4}} = \sqrt[4]{\frac{2\cdot3^4x^2}{2^4}} = \frac{3\sqrt[4]{2x^2}}{2}$$

  • $\begingroup$ In your second approach you have $\sqrt[4]{\frac{2\cdot3^4x^2}{2^4}}$. Why is it not $x^6$ in the numerator there? $\endgroup$
    – Doug Fir
    Jan 6, 2019 at 21:24
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    $\begingroup$ I simplified $\frac{x^6}{x^4}$ first. $\endgroup$
    – KM101
    Jan 6, 2019 at 21:25
  • $\begingroup$ In your first answer, your numerator goes from $3\sqrt[4]{2x^4x^2}$ to $3\vert x\vert\sqrt[4]{2x^2}$. I see the benefit of pulling an x out in front of the radical but cannot see how you did that? Would it be possible to expand on that part if you have a minute? $\endgroup$
    – Doug Fir
    Jan 6, 2019 at 21:40
  • $\begingroup$ Sure. $\sqrt[4]{x^4} = \vert x\vert$, like how $\sqrt{x^2} = \vert x\vert$, $\sqrt[3]{x^3} = x$, etc. (As another point, note the use of absolute values when the index is even because the answer is always positive. For an odd index, as in cube roots, sign is preserved, so no absolute value is used.) $\endgroup$
    – KM101
    Jan 6, 2019 at 21:47
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    $\begingroup$ A good way of thinking about it: $$\sqrt[4]{2x^6} = \sqrt[4]{2x^2}\cdot\sqrt[4]{x^4}$$ The first quartic root can’t be simplified, and it is therefore left as it is. The second simplifies to $\vert x\vert$. So, the $x$ comes from $x^4$. The $2x^2$ stays inside the radical. (I think your confusion is coming from the $2$ in front of the $x^4$. Bringing out $\sqrt[4]{2}$ would literally be... $\sqrt[4]{2}$ again, so you don’t bring it out as it won’t simplify anything. The exponent of the base must be greater than $4$ for it to be brought out.) $\endgroup$
    – KM101
    Jan 6, 2019 at 22:00

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