steps: √(2x+1) = 𝒊 square both sides 2x+1 = -1 subtract 1 2x = -2 divide by 2 x = -1 checking: substitute x = -1 √(2∙[-1]+1) = 𝒊 √(-2+1) = 𝒊 √(-1) = 𝒊 𝒊 = 𝒊

nice

i just stumbled to this 10 mins ago when trying to solve random maths, solved it myself, then yt showed this to me?

Does ³√x = −1 have solutions over ℝ and does it have solutions over ℂ? Are they the same?

@Ninja20704

4 ай бұрын

Yes there is real solutions, i mean simply cbrt(-1)=-1

@allozovsky

4 ай бұрын

@@Ninja20704 And what about solutions over ℂ? Is x = −1 also a solution of ³√x = −1 over the field of complex numbers with the definition of complex roots? The notation for the cube root keeps the same.

@bolrod_4526

4 ай бұрын

​​@@allozovsky There is only one solution to solve a cube root that is equal to a real number since the parity of the root allows for negative R , I do not know how expanding our domain to the complex domain allows for more solutions but maybe I haven't gone far enough in my studies

@sohailansari07289

4 ай бұрын

The possible solutions for x is as follows: For x€R, x=-1 For x€C, x= e^[3πi(2n+1)], for n€Z How you obtain this is basically convert -1 into e^iØ(Complex form), take natural log on both sides and then exponentiate them to e.

@allozovsky

4 ай бұрын

@@bolrod_4526 The principal cube root of −1 is normally defined to be equal to ³√(−1) = (1 + 𝒊√3)/2, so under this definition x = −1 is no longer a solution to ³√x = −1 over ℂ, which looks pretty strange, since we have lost a solution after expanding the domain.

Does it help us solve √(2x+1) = −𝒊 though?

@Logay495

4 ай бұрын

It's absolvable

@allozovsky

4 ай бұрын

@@Logay495 Solving irrational equations over the field of complex numbers is not a very good idea in the first place, since complex roots are essentially multivalued and the notation ⁿ√z is ambiguous. We have two choices: keep the expression single-valued and define it to be equal to a so-called principal root (which can be done in different ways), or treat it as multivalued and redefine what a solution to an equation is (when it contains multivalued functions). Both ways are possible. In the latter case the above equation has a solution (the same as in the video).

@Logay495

4 ай бұрын

@@allozovsky thank you so much for increasing my knowledge

A New series from beginner (10th standard) to Olympiad level

x=-1

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