Imagine x^1/3 as a different variable and solve for it
@OniDevv
2 ай бұрын
let y=x^1/3 we have the exact question in our highschool book and this question might come in our exam 😭 edit: my exam is in 20 days and i am fucked
@allozovsky2 ай бұрын
But which roots/powers are we considering? The real-valued ones or the principal values of the complex functions? For the principal value of the complex cube root the equation x¹ᐟ³ = ³√x = −2 has no solutions (neither real, nor complex). While it is true that (−2)³ = −8, nonetheless the _principal_ complex cube root of −8 = 2³·exp(𝒊π) is ³√(−8) = 2·exp(𝒊π/3) = 1 + 𝒊√3, not ³√(−8) = −2 (which is the real-valued cube root). So this solution is somehow ambiguous.
@nicolastorres1472 ай бұрын
-8 is actually an extraneous root if taking the main branch in C
@allozovsky
2 ай бұрын
Exactly. Though 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 a so-called _"principal"_ value may be chosen arbitrarily by (as Wikipedia puts it) "requiring the argument arg(z) to belong to one, _conveniently_ selected turn", say, −π < φ ≤ π, or 0 ≤ φ < 2π, or some other half-open interval of length 2π. Yes, normally it is −π < φ ≤ π and most tools use this definition, but it's only one of an (infinite) range of options and crossing our other ones is not "fair".
@allozovsky
2 ай бұрын
The equation ³√x = −2 having solutions over ℝ and not having them over ℂ (for the principal root definition) is a pretty strange situation anyway. Some math books allow complex roots to be maltivalued and generalize _equations_ f(x)= 0 to _inclusions_ f(x) ∋ 0, so under this _generalized_ (re)definition, ³√x = −2 gets back its solutions over ℂ, since ³√(−8) = {−2; 1 ± 𝒊√3} ∋ −2, which looks much more acceptable, as now we do not lose our roots when switching from ℝ to ℂ.
@allozovsky
2 ай бұрын
Steve claimed in a recent video that √1 = 1 only, not ±1 (and even gave ³√8 as an example to support his claim), while at the same time in an older video he evaluated √𝒊 to both √𝒊 = (1 + 𝒊)/√2 and √𝒊 = (1 − 𝒊)/√2 no problem, starting the video with "as we all know, √(−1) = 𝒊", and ending it with "there are two answers for √𝒊".
@nabilal.rawahi98472 ай бұрын
Nice
@rudrapatel68902 ай бұрын
Solved in 5 seconds just by seeing it 😎😎😎
@lzuluaga60642 ай бұрын
Haz x^1/3 =u.
@user-qc9cd5iz3l2 ай бұрын
±6.75??
@user-qc9cd5iz3l
2 ай бұрын
Where did I go wrong? I was doing it in my head really fast so the chances of getting wrong answer is high Let x^⅓ be ‘y’, I didn't even do this step in my head lol y² - y - 6 = 0 y² - y + ¼ = 6.25 (y - ½)² = 6.25 y = (0.5 ± √6.25) x = y³ √6.25 = 2.5 y = (-2,3) x = (-8,27)
@og_jeremie2 ай бұрын
let y=x^1/3 solve for y then substitute back for x 👍🏾
Пікірлер: 14
Imagine x^1/3 as a different variable and solve for it
@OniDevv
2 ай бұрын
let y=x^1/3 we have the exact question in our highschool book and this question might come in our exam 😭 edit: my exam is in 20 days and i am fucked
But which roots/powers are we considering? The real-valued ones or the principal values of the complex functions? For the principal value of the complex cube root the equation x¹ᐟ³ = ³√x = −2 has no solutions (neither real, nor complex). While it is true that (−2)³ = −8, nonetheless the _principal_ complex cube root of −8 = 2³·exp(𝒊π) is ³√(−8) = 2·exp(𝒊π/3) = 1 + 𝒊√3, not ³√(−8) = −2 (which is the real-valued cube root). So this solution is somehow ambiguous.
-8 is actually an extraneous root if taking the main branch in C
@allozovsky
2 ай бұрын
Exactly. Though 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 a so-called _"principal"_ value may be chosen arbitrarily by (as Wikipedia puts it) "requiring the argument arg(z) to belong to one, _conveniently_ selected turn", say, −π < φ ≤ π, or 0 ≤ φ < 2π, or some other half-open interval of length 2π. Yes, normally it is −π < φ ≤ π and most tools use this definition, but it's only one of an (infinite) range of options and crossing our other ones is not "fair".
@allozovsky
2 ай бұрын
The equation ³√x = −2 having solutions over ℝ and not having them over ℂ (for the principal root definition) is a pretty strange situation anyway. Some math books allow complex roots to be maltivalued and generalize _equations_ f(x)= 0 to _inclusions_ f(x) ∋ 0, so under this _generalized_ (re)definition, ³√x = −2 gets back its solutions over ℂ, since ³√(−8) = {−2; 1 ± 𝒊√3} ∋ −2, which looks much more acceptable, as now we do not lose our roots when switching from ℝ to ℂ.
@allozovsky
2 ай бұрын
Steve claimed in a recent video that √1 = 1 only, not ±1 (and even gave ³√8 as an example to support his claim), while at the same time in an older video he evaluated √𝒊 to both √𝒊 = (1 + 𝒊)/√2 and √𝒊 = (1 − 𝒊)/√2 no problem, starting the video with "as we all know, √(−1) = 𝒊", and ending it with "there are two answers for √𝒊".
Nice
Solved in 5 seconds just by seeing it 😎😎😎
Haz x^1/3 =u.
±6.75??
@user-qc9cd5iz3l
2 ай бұрын
Where did I go wrong? I was doing it in my head really fast so the chances of getting wrong answer is high Let x^⅓ be ‘y’, I didn't even do this step in my head lol y² - y - 6 = 0 y² - y + ¼ = 6.25 (y - ½)² = 6.25 y = (0.5 ± √6.25) x = y³ √6.25 = 2.5 y = (-2,3) x = (-8,27)
let y=x^1/3 solve for y then substitute back for x 👍🏾
Sir please pin me