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https://www.wolframalpha.com/input/?i=%5Csqrt%7B3x-4%7D%2B%5Csqrt%5B3%5D%7B5-3x%7D%3D1

This surely has the third solution as x= 13/3.

But:

https://www.wolframalpha.com/input/?i=x%3D%3D13%2F3%2C+%5Csqrt%7B3x-4%7D%2B%5Csqrt%5B3%5D%7B5-3x%7D-1

Why Wolfram fails to get it correct?

Brady Zeng
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  • You do know that $-1$ isn't the only value associated with $\sqrt[3]{-1}$? In fact, when the latter is written, it almost always doesn't mean $-1$. – Rushabh Mehta Sep 14 '19 at 19:39
  • I expect it is confused as to which branch of the cube root you intend. I specified the real-valued root and it found the third root with no problem. – lulu Sep 14 '19 at 19:39
  • yes $$\sqrt[3]{-1}$$ means three things but it surely can mean one thing of the three and satisfies the equation, but why wolfram still fails to get it? – Brady Zeng Sep 14 '19 at 19:50
  • @BradyZeng because it works with principal root, unless instructed otherwise. Although if you use 1/3 instead of sqrt[3], it "magically" gives you the option of using the real root. – Momo Sep 14 '19 at 19:52

2 Answers2

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Because it assumes the principal cube root. If you add the assumption that the root is the real root, it gives you all the solutions like here.

Momo
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The Wolfram Language function Power[] (documentation) gives the principal root, meaning the one with the least complex angle. For example, \sqrt[3]{-8}, which gives $1 + \mathrm{i}\sqrt{3}$, having angle $\pi/3$, instead of $-2$, having angle $\pi$. We can see that the complex angle of $\sqrt[3]{5-3x}$ changes between 1.6 and 1.7.

WA: plot (5 - 3 x)^(1/3) and Arg((5 - 3 x)^(1/3)) for 0<=x<=5

in exactly the same way the result of Power[] does.

WA: plot Power[5 - 3 x,1/3] and Arg(Power[5 - 3 x,1/3]) for 0<=x<=5

Methods to avoid this

  • Use a function that always returns the real root, like CubeRoot[] (doc). \sqrt{3x-4}+CubeRoot[5-3x]=1. Also cbrt() is a WA shortcut for this function.
  • Use a form for which WA explicitly mentions that it is assuming a root, then tell it which assumption you actually want. \sqrt{3x-4}+(5-3x)^(1/3)=1 , then click "the real-valued root instead".
  • More generally than using CubeRoot[], there is surd() (docs), which also gives the real root, when there is one. So here, we would use \sqrt{3x-4}+surd(5-3x,3)=1
Eric Towers
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