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How to show that this equation $ \sqrt[5]{20-2x} + \sqrt[5]{7-x}+\sqrt[5]{3x+5}=2$ have 3 solutions 9, -6 and -25

Wolframsays the equation have no solutions !

Pascal
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  • Your input may be problematic. Principal root and Real valued root are two different things generally. – lone student Mar 10 '21 at 13:02
  • Also, usually when you input $\text{(x)^(1/5)}$ wolfram realises that you might want to say the real root instead of the principal value of the complex root, but, if you start inputing latex code, wolfram dedicates its resurces to interpreting what you're writing, rather than assessing alternative interpretations of your input. –  Mar 10 '21 at 13:09

2 Answers2

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If $\sqrt[5]{20-2x}= a$ etc., $a+b+c=2$

$$a^5+b^5+c^5=32=(a+b+c)^5$$

Use How to factor $(a +b+c) ^5 -(a^5+ b^5 + c^5)$? to discover that $(a+b)(b+c)(c+a)=0$ should give three solutions.

lab bhattacharjee
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  • Based on real-valued root, not principal. – lone student Mar 10 '21 at 13:05
  • @lonestudent, Please elaborate – lab bhattacharjee Mar 10 '21 at 13:09
  • Please do not view this as a negative comment.. I mean, if we take the principal root under the odd degree radical, then actually there is no real solution, right? – lone student Mar 10 '21 at 13:14
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    @lonestudent, No no, its fine. I just wanted to learn your point. Please feel free to comment for your query/doubt – lab bhattacharjee Mar 10 '21 at 13:19
  • Ah ok https://www.wolframalpha.com/input/?i=%2820-2x%29%5E%281%2F5%29+%2B+%287-x%29%5E%281%2F5%29%2B%283x%2B5%29%5E%281%2F5%29%3D2&assumption=%22%5E%22+-%3E+%22Real%22 – Pascal Mar 10 '21 at 13:21
  • Thank you @lab bhattacharjee – Pascal Mar 10 '21 at 13:30
  • What should we do when we actually see such an equation? For example, let's say a teacher gave homework. My teacher, "what root use do you allow?" Is it the real-valued root or is it the principal root? What should we base on when solving the equation? Or is that unimportant? Should we calculate directly on real valued roots?Thank you. – lone student Mar 10 '21 at 14:01
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The claim that $9,-6,-25$ are the solutions of the equation does not hold.

Substituting for example $x=9$ into the equation one obtains: $$ (2)^\frac15+(-2)^\frac15+(2^5)^\frac15\ne2 $$ because $$ (2)^\frac15+(-2)^\frac15\ne0 $$ for any pair of the fifth order roots of $2$ and $-2$ belonging to the same branch.

user
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  • For me @user $(-2)^\frac15=-2^\frac15$ – Pascal Mar 10 '21 at 18:41
  • The two roots reside in different branches of $z^{1/5}$. – user Mar 10 '21 at 18:44
  • I was taught that $ x \to x^{2n + 1} $ is a bijection $\mathbb{R}\to \mathbb{R}$, its reciprocal function is $ x \to x ^ {\frac 1 {2n + 1}} = \sqrt [2n + 1] {x} $ and $\forall x\in \mathbb R, \sqrt [2n + 1] {-x}=-\sqrt [2n + 1] {x}$ It's a shock for me to tell otherwise – Pascal Mar 10 '21 at 18:54
  • I believe that the real definition of the function is out-of-date and certainly is not used by default by WA: https://www.wolframalpha.com/input/?i=+%5Csqrt%5B5%5D%7B-2%7D. – user Mar 10 '21 at 19:12
  • I take it by this discussion that such branches do not "mix and match" as it were? Certainly the original equation does not show any particular limitations on the available branches... – abiessu Mar 11 '21 at 06:05
  • In my opinion if an equation contains several instances of a multi-valued function, one should assume that all the instances refer to the same branch of the function. – user Mar 11 '21 at 07:34