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The equation is:

x  Log[x] - 3/2 x - 1/(2 x) + 2 == 0

Its root is just one, which is 1. Why does the output repeatedly write out 1 multiple times?

Solve[x  Log[x] - 3/2 x - 1/(2 x) + 2 == 0, x, Reals]

get

{{x -> 1}, {x -> 1}, {x -> 1}}
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  • $\begingroup$ Hmmm... weird... never seen that. ($+1$). I played around with real-values (-1.5 rather than -3/2), but it didn't make a difference. $\endgroup$ Apr 21 at 0:23
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    $\begingroup$ Please read the documentation for Solve. When a single variable is specified and a particular root of an equation has multiplicity greater than one, Solve gives several copies of the corresponding solution. $\endgroup$
    – Domen
    Apr 21 at 6:58

1 Answer 1

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Have a look at

Series[x Log[x] - 3/2 x - 1/(2 x) + 2, {x, 1, 4}]

$$\frac{1}{3} (x-1)^3-\frac{5}{12} (x-1)^4+O\left((x-1)^5\right)$$

Its of course the trick to generate functions with multiple zeros, to subtract the leading powers of a series

  Series[x Log[x] , {x, 1, 4}]

$$(x-1)+\frac{1}{2} (x-1)^2-\frac{1}{6} (x-1)^3+\frac{1}{12} (x-1)^4+O\left((x-1)^5\right)$$

Series[-3/2 x - 1/(2 x) + 2, {x, 1, 4}] $$-(x-1)-\frac{1}{2} (x-1)^2+\frac{1}{2} (x-1)^3-\frac{1}{2} (x-1)^4+O\left((x-1)^5\right)$$

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