# Get a real-valued root?

Posted 1 year ago
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|   As you can see,I want to find the real-valued root in Mathematica, what should i do? I tried so hard to search it up online, but in vein. Answer
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Posted 1 year ago
 Maybe try Surd instead of Sqrt. Answer
Posted 1 year ago
 Where do you mean I should try Surd with? I mean I use 1/3 power, isn't that the same? Answer
Posted 1 year ago
 Another way, using CubeRoot CubeRoot[7 + Sqrt] + CubeRoot[7 - Sqrt] // N Answer
Posted 1 year ago
 Thanks, it works, but is that really the only way to solve it? What if there are something like 1/5 power and the answer still come up with imaginary number? Answer
Posted 1 year ago
 Then you can use Surd, as Henrik suggested Surd[7 + Sqrt, 5] + Surd[7 - Sqrt, 5] // N CubeRoot is just a special form of Surd. The following will give the same result as if CubeRoot was used, in case of power 1/3: Surd[7 + Sqrt, 3] + Surd[7 - Sqrt, 3] // N Answer
Posted 1 year ago
 The 1/5 power in Mathematica is defined as x^(1/5) == Exp[(1/5) Log[x]]. When x is negative, the principal value of the logarithm has I Pi as imaginary part, so that In:= (-1)^(1/5) == Exp[(1/5) Log[-1]] Out= True and In:= Exp[(1/5) Log[-1]] // N Out= 0.809017 + 0.587785 I This issue is an endless source of confusion and frustration for beginners, which had been raised in a real-centered theory of roots. Answer
Posted 1 year ago
 Thanks everyone for helping. Answer