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Given $y^3=-x^2$ I thought the solution was $y=-x^{(2/3)}$ but as I see from the solution of my prof He wrote $y=-|x|^{(2/3)}$.

I don’t understand the absolute value since we’re extracting an odd root, what am I missing?

Eric Wofsey
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Andrew Carrol
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4 Answers4

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Both expressions are actually equivalent. Note: $y = −x^{(2/3)}$ $\implies y = -(\sqrt[3]{x})^2$

Taking the absolute value of x doesn't change the function as we square the term anyway.

Raiyan Chowdhury
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if $x$ is a solution, the $|x|$ is a solution as well since $|x|^{2/3}= (|x|^2)^{1/3} = (x^2)^{1/3}=x^{2/3}$

alphaomega
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You are correct.

$$y = (-x^2)^{1/3} = (-1)^{1/3}(x^2)^{1/3}=-(x^2)^{1/3}=-x^{2/3}$$

The absolute value was brought in to make sure what was already sure, not really necessary.

Narasimham
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  • Now I think you were more right and I deleted my comment :) I don't know for sure :) it's just confusing. But I think in real analysis when they write $x^a$ they just mean that this implies $x \ge 0$ And I am pretty sure that's why the professor gave this answer (he just made sure what he raises to power is non-negative). – peter.petrov Jul 25 '20 at 22:34
  • :) How about CAS evaluation hierarchy giving for $ (-8^2)^{(1/3.)} [\ne - 4]$ , first complex root $...(-8^2)^{(1/3.)} == (2.+ 3.4641016151377535 i)$ as True, and its conjugate $(-8^2)^{(1/3.) }== (2.- 3.4641016151377535 I) $ as False ! – Narasimham Jul 26 '20 at 05:59
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I had a similar confusion recently.

See:
How exactly is the function $x^a$ defined?

So the key thing to realize here is this:

NOTE: In real analysis when they use/write $x^a$ (for any $a$ real) they usually define it just for $x \ge 0$. This is to avoid all sorts of complications and ambiguities.

And when I say for any $a$ real I mean the power $a$ can be positive, negative, zero, rational or irrational.

So the way you have written your answer it implies $x \ge 0$ which is not necessarily true.

And it's in that sense (as others said) that the professor is more correct.

Important here is to realize that the NOTE kicks in when you're in real analysis.

peter.petrov
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  • So it’s a matter of definitions, i had this feeling but can’t find a real answer. Thanks for the material :) – Andrew Carrol Jul 25 '20 at 22:47
  • @AndrewCarrol Yeah I think so. In high school I would give your answer by the same logic as yours. But in real analysis in a university when they write $x^a$ they imply $x \ge 0$ At least that's my current understanding. – peter.petrov Jul 25 '20 at 22:49