Multiply numerator and denominator of the complex solutions with ³√2 to rationalize them. You get ± ³√2 / 2 ± ³√2 ⋅ i ⋅ √3 / 2 = ± (³√2 ± i ⋅ ⁶√108)/2. Or simply ³√2 / 2 ⋅ (±1 ± i ⋅ √3).
show the roots as six equally spaced points around the circumference of a circle radius (2^(1/3)) on an Argand diagram. The two real roots can be found by inspection after the first step. The other 4 can be identified in polar coordinates, almost by inspection.
x⁶=4 so |x|=2^(1/3) For an angle u we have x=2^(1/3)×e^(iu) So e(i6u)=e^(i2k×pi) for an integer k So u=k×pi/3 and by taking for k the first 6 positives integer (including 0) we obtain: x1=2^(1/3) x2=2^(1/3)×e^(i×pi/3) x3=2^(1/3)×e^(i×2pi/3) x4=2^(1/3)×e^(i×pi)=-2^(1/3) x5=2^(1/3)×e^(i×4pi/3) x6=2^(1/3)×e^(i×5pi/3) That's such an easy question, for Harvard entrance (I am not american so I don't know how difficult it is supposed to be).
This is the type of problem that a college maths professor will only give you 5 minutes to do on a timed exam. Takes this guy 16 minutes without any mistakes. Hilarious.
lol, I did it in about 10 secs by inspection. first root is 2^(1/3) other five you get from taking that magnitude and multiplying by exp((i*pi/3))^n -- n goes from 1 to 5
To solve the equation x^6 - 4 = 0, we can start by adding 4 to both sides, which gives us: x^6 = 4 Next, we can take the sixth root of both sides, which gives us: x = ∛∛2 or x = ±∛∛2 So, the solutions to the equation are: x = ∛∛2 and x = -∛∛2 Note that ∛∛2 is the sixth root of 2, which is approximately 1.26.
This reminds me of a math quiz bowl question I got in high school (a buzzer speed game like Jeopardy, no paper or pencil much less anything else allowed). "Simplify ⁶√343." I instantly recognized 343=7³. I buzzed in and said "plus or minus the square root of seven", knowing full well that there were four complex roots, but not being prepared to rattle them off quickly, and hoping that wasn't expected. It took them a while to accept my answer - I seriously think they had to verify that negative root 7 was also an answer. The other four roots are "plus or minus the square root of seven, plus or minus i times the square root of 21, all divided by two".
You should put your answer in standard form. The magnitude of the 6 solutions is ³√2 times the 6 sixth roots of 1. So it's best to write: x = ±³√2, ±³√2*(1+i*√3)/2, ±³√2*(1-i*√3)/2
Off the top of my head: Real solutions: +/- 3rd root of 2. Polynomial division by (x² - 3rd root of 4) should* yield a biquadric equation that can be turned into a regular quadratic equation by substitution yielding two pairs of complex solutions => in total 6 solutions, 2 real, 4 complex. *have yet to check whether that assumption is true.
No quite. If you apply the third binomial formula, you get x⁶ − 4 = (x³ − 2) ⋅ (x³ + 2) = 0. This gives you x₁ = ³√2 and x₂ = −³√2 as the real solutions. If you then factor out (x − ³√2) for the first and (x + ³√2) for the second cubic function (x³ − 2 = 0 and x³ + 2 = 0, respectively), and solve the resulting quadratic formulas, you get the four complex solutions.
As an entrance exam, assuming the takers have only the minimally required highschool algebra, this is a very doable but long problem.. anybody with more advanced math experience will be able to do this in their sleep.
Interesting, but long winded. Why not just use De Moivre's theorem? Also, the question as posed does not state that the solution has to be real, but many people would take that as read. In that case the only extra points would be for spotting the negative solution. The question is not precise enough.
Because using rote learned algorithms is not actually conceptualization which is what an entrance exam is ultimately testing. Or ability to derive it on your own.
Problems like this required advanced techniques before numerical computing came along 50-60 years ago. In 15 sec, I get x= 1.259921049894873 from a scientific calculator. Some will call into question why Harvard still holds onto pre-computational challenges like this? Tradition?
Actually, this problem has been simple much longer, since about year 1750. Nothing more advanced than Euler's Formula required. In my head, I knew the answer was cube-root 2 times each of the 6 complex roots of unity.
@@KennethDuda OK, but if I need to solve this for a practical application, I still need numerics for the cube root of 2. I'm going to consult a calculator. So why not just enter the 6th root of 4? I still don't see the point of Euler's formula if I can calculate y√x.
@@pbierre If working on an engineering problem, especially if it's to do with vibrations/oscillations, one might need the complex roots as well. Also in acoustics, signal processing, classical waves, quantum mechanics they can be important...They might correspond to different phases of oscillations, different spacings of elements in a structure, different times an aftershock might come after an earthquake, different particles or particle states, myriads of things. But hopefully by the time anyone started working on any of that stuff they'd know all about the polar form of complex numbers and Euler's formula and could go straight to the answer instead of doing all that tedious algebra.
@@pbierre Yep, what zetacrucis681 said --- the problem has six solutions, and they might all have practical relevance depending on why we're trying to solve x^6 - 4 = 0. So we might need all six. Typically, in mathematics, "solving" something means reducing it to a form "x = " where the expression involves well-known constants and functions. Thus, x^6 - 4 = 0 is not solved, but x = (2^(1/3)) * e^(k * pi * i / 3) for k = 0,1,2,3,4,5 would be considered solved. Thus you don't "need numerics" to get to the solution of this problem. If you want a numerical approximation for x, then we can get out our calculators etc., but it might be worth keeping in mind that once you enter 4^(1/6) in your calculator, you actually no longer have the exact right answer because there is no precise (finite) decimal representation of that number. What is the point of all this? That's a question for philosophers, not mathematicians!
There are five other solutions. Take the 6 complex roots of unity, and multiple each by the cube-root of 2, and those are the six solutions. (Note that -1 and 1 are two of the six complex roots of unity, yielding the real solutions 2^(1/3) and -2^(1/3), but there are four other roots of unity, yielding four more solutions. One of those is x = 2^(1/3) * (cos pi/3 + i * sin pi/3).)
At 7:15, you should have used a different variable from x since x was already the variable in the original equation! Also, I do not like mixing radical and exponential forms in the same expression.
"Higher Mathematics", this is totally obvious, it's just each of the 6 complex roots of unity, times the cube-root of 2. I could do that in my head. Why the long video? Why difference of squares and sum of cubes formula? None of this is needed. You are making an easy problem hard.
@@daangueto Yes, the six solutions are 2^(1/3) * e^(k * pi * i / 3) for k = 0, 1, 2, 3, 4, 5. Sorry, that was what I meant by "each of the 6 complex roots of unity, times the cube-root of 2".
On y arrive aussi en recherchant les n racines d’un nombre complexe z: z^n=r.e^i.2kpi/n avec k E {0;n-1} avec r=4 et n=6 Ce qui donne: z^6=4.e^i 2kpi/6 avec k E {0,5} Cela oblige à repasser à la forme a+ib: z=r sin téta + i r cos téta Avec r=4^1/6 et téta=pi/3 et ses multiples. Il est vrai que ça tombe bien car, dans ce cas, les 2kpi/6 tombent sur des valeurs dont on connais les sin et cos. J’accepte volontiers les critiques 😢
X^6 - 4 = 0 X^6 = 4 X = 6√4 ( 6 here is used to show the order of the surd) X = 6√2^2 X= 3√2 ( again ,3 here is the order of the surd) No need of all that complicated stuff
@@horrifichalo8430 why should there be 6 solutions if any positive integer exponentiation multiple of two is a parabolic function and has only 2 points of intersection with a linear function?
So würde ich lösen: x⁶ − 4 = 0 3. binomische Formal anwenden: (x³ − 2) ⋅ (x³ + 2) = 0 Satz vom Nullprodukt ergibt schon mal: x₁ = ³√2 ∨ x₂ = −³√2 Diese Lösungen kann man jeweils aus den kubischen Funktionen ausklammern: (x − ³√2) ⋅ (x² + ³√2 x + ³√4) ⋅ (x + ³√2) ⋅ (x² − ³√2 x + ³√4) = 0 Und nun kann man die beiden quadratischen Funktionen (wieder Satz vom Nullprodukt) mit der pq-Formel lösen, um die vier komplexen Lösungen zu erhalten: x₃,₄ = −³√2 / 2 ± √(³√4 / 4 − ³√4) = −³√2 / 2 ± ³√2 ⋅ i ⋅ √3 / 2 = −³√2 / 2 ± ⁶√108 ⋅ i / 2 x₃ = (−³√2 − ⁶√108 ⋅ i) / 2 ∨ x₄ = (−³√2 + ⁶√108 ⋅ i) / 2 x₅,₆ = ³√2 / 2 ± √(³√4 / 4 − ³√4) = ³√2 / 2 ± ³√2 ⋅ i ⋅ √3 / 2 = ³√2 / 2 ± ⁶√108 ⋅ i / 2 x₅ = (³√2 − ⁶√108 ⋅ i) / 2 ∨ x₆ = (³√2 + ⁶√108 ⋅ i) / 2 Dann kann man noch die Lösungsmenge mit den sechs Lösungen angeben, aber das schenke ich mir hier mal. Genau wie die Probe.
Well, I got cube root of two without pencil and paper. X^6-4=0 6 log(x) = log(4) Use base 2 logs. 6 log(x)=2 Log(x)=1/3 X= 2^1/3. Plus or minus since we’re raising x to an even power in the original equation.
@@Nikioko please share. I found plus and minus cube root of 2. Is there another number that equals 4 when raised to the 6th power? I would like to learn. The square root of minus one mystifies me.
@@johnnyragadoo2414 x⁶ − 4 = 0 Third binomial formula a² − b² = (a + b) ⋅ (a − b): (x³ − 2) ⋅ (x³ + 2) = 0 Rule of the zero product: The product is zero when at least one of the factors is zero, so: x³ − 2 = 0 or x³ + 2 = 0 And thus: x³ = 2 or x³ = −2 By taking the third root, we get: x₁ = ³√2 or x₂ = −³√2 Let's take the first factor of (x³ − 2) ⋅ (x³ + 2) = 0: x³ − 2 = 0 for x = ³√2, obviously. So let's factor out x − ³√2: (x − ³√2) ⋅ (x² + ³√2 x + ³√4) = 0 Again, rule of the zero product. We know that with x = ³√2, the first factor becomes zero. So let's make the second factor zero: x² + ³√2 x + ³√4 = 0 Now you can solve the quadratic formula the way you prefer: x₃,₄ = −³√2 / 2 ± √(³√4 / 4 − ³√4) = −³√2 / 2 ± √(−3 ⋅ ³√4 / 4) = −³√2 / 2 ± √−3 ⋅ √(³√4 / 4) = −³√2 / 2 ± i ⋅ √3 ⋅ (³√2 / 2) = ³√2 / 2 ⋅ (−1 ± i ⋅ √3) Now you have two complex solutions. If you factor out x + ³√2 on x³ − 2 = 0 and solve the quadratic formula again, you get the other two complex solutions: x₅,₆ = ³√2 / 2 ⋅ (1 ± i ⋅ √3) So, overall, you get the two real solutions x₁,₂ = ±³√2 and the four complex solutions x₃,₄,₅,₆ = ³√2 / 2 ⋅ (±1 ± i ⋅ √3).
Si a et b sont des nbres complexes, avec, en polaires, a =(|a|,arg(a)) et arg(a) € [0;2pi[, et b = x+i.y, on a, en polaires, a^^b = ( exp(x.ln(r) - y.arg(a)) , x.arg(a) + y.ln(r) ), ceci est vrai ¥a et ¥b (pourvu que arg(a) € [0, 2pi[). Ici a^^6 = ( exp(6.ln(|a|) , 6.arg(a) ) et 4 = ( 4 , 0 ) (tout en polaires) ==> exp(6.ln(r)) = 4 et 6.arg(a) = 0 + 2k.pi. D'où les 6 racines. Rq ln est le logaritme réel.
Looks weird. Cause no one needs to take the answer with complex numbers. And every time all solutions are based on a simple rule - "D" does not equal "- value". So, X 3456 no need here. In the real world it calls - over engineering
Multiply numerator and denominator of the complex solutions with ³√2 to rationalize them. You get ± ³√2 / 2 ± ³√2 ⋅ i ⋅ √3 / 2 = ± (³√2 ± i ⋅ ⁶√108)/2.
Or simply ³√2 / 2 ⋅ (±1 ± i ⋅ √3).
show the roots as six equally spaced points around the circumference of a circle radius (2^(1/3)) on an Argand diagram. The two real roots can be found by inspection after the first step. The other 4 can be identified in polar coordinates, almost by inspection.
x = 2^1/3* cis(N*pi/3) --- by the Nth roots of unity scaled by 2^1/3
Fully agree!!!
x⁶=4 so |x|=2^(1/3)
For an angle u we have
x=2^(1/3)×e^(iu)
So e(i6u)=e^(i2k×pi) for an integer k
So u=k×pi/3 and by taking for k the first 6 positives integer (including 0) we obtain:
x1=2^(1/3)
x2=2^(1/3)×e^(i×pi/3)
x3=2^(1/3)×e^(i×2pi/3)
x4=2^(1/3)×e^(i×pi)=-2^(1/3)
x5=2^(1/3)×e^(i×4pi/3)
x6=2^(1/3)×e^(i×5pi/3)
That's such an easy question, for Harvard entrance (I am not american so I don't know how difficult it is supposed to be).
This gets a lot easier if you let c = 2^(1/3). Then just remember to resubstitute the answer as you get close to done :)
I love this and this old engineer has learned a lot by reading the comments. I wish we had internet while I was in school. My thanks go to all.
x^6-4=0 x= ±Surd[2,3] x=±0.5Cbrt[2]±0.5Surd[108,6]i=(±Cbrt[2]±Surd[108,6])/2= ± 1/2^(2/3) ±( sqrt(3))/(2^(2/3))i
16 minutes? This should only take 30 seconds if we start with Euler's identity
Sure that you get all 6 solutions in 30 seconds?
This is the type of problem that a college maths professor will only give you 5 minutes to do on a timed exam. Takes this guy 16 minutes without any mistakes. Hilarious.
Because hes explaining it along the way ? I dont know why there always have to be someone like you who need to jerk with his huge ego
@@Nikiokoabsolutely. Write the exponential as i ( 0 + 2 k pi). Divide by 6 and see how many k’s you need before you reach 2 pi
lol, I did it in about 10 secs by inspection. first root is 2^(1/3) other five you get from taking that magnitude and multiplying by exp((i*pi/3))^n -- n goes from 1 to 5
first thought, if x^6=4, then x^3=2; and x=+/-(3√2)
Simply take log both sides with any valid base,
Or, write the equation as x^6=4, imply x=4^(1/6)=2^(1/3).
Yes. This is one solution. but there could be 5 more.
To solve the equation x^6 - 4 = 0, we can start by adding 4 to both sides, which gives us:
x^6 = 4
Next, we can take the sixth root of both sides, which gives us:
x = ∛∛2
or
x = ±∛∛2
So, the solutions to the equation are:
x = ∛∛2 and x = -∛∛2
Note that ∛∛2 is the sixth root of 2, which is approximately 1.26.
This reminds me of a math quiz bowl question I got in high school (a buzzer speed game like Jeopardy, no paper or pencil much less anything else allowed). "Simplify ⁶√343." I instantly recognized 343=7³. I buzzed in and said "plus or minus the square root of seven", knowing full well that there were four complex roots, but not being prepared to rattle them off quickly, and hoping that wasn't expected. It took them a while to accept my answer - I seriously think they had to verify that negative root 7 was also an answer. The other four roots are "plus or minus the square root of seven, plus or minus i times the square root of 21, all divided by two".
Move 4 to the other side, take the 6th root of both sides, x=6th root of 4
Yea, that was also my solution. You can solve this in 30 seconds. We should go to harvard i think ;)
Which is ³√2 and only one of six solutions.
Obv + and - 6th root bc it’s an even root
Is it right to keep the answer in the root form? Feels like the question became too easy when we do like this@@Nikioko
Your solution isn't simplified though.
I guess de Moivre's equation was not allowed at Harvard.
You should put your answer in standard form.
The magnitude of the 6 solutions is ³√2 times the 6 sixth roots of 1. So it's best to write:
x = ±³√2, ±³√2*(1+i*√3)/2, ±³√2*(1-i*√3)/2
Just use roots of unity, way easier than all this nonsense.
We need the 6 solutions of the problem, not only 1
@@daangueto Yea, and that’s the point of roots of unity…
4 ^ (1/6) * exp ( 2 π i r /6)
Herein r = 0, 1, 2, 3, 4 , 5
Off the top of my head: Real solutions: +/- 3rd root of 2. Polynomial division by (x² - 3rd root of 4) should* yield a biquadric equation that can be turned into a regular quadratic equation by substitution yielding two pairs of complex solutions => in total 6 solutions, 2 real, 4 complex.
*have yet to check whether that assumption is true.
No quite. If you apply the third binomial formula, you get x⁶ − 4 = (x³ − 2) ⋅ (x³ + 2) = 0.
This gives you x₁ = ³√2 and x₂ = −³√2 as the real solutions.
If you then factor out (x − ³√2) for the first and (x + ³√2) for the second cubic function (x³ − 2 = 0 and x³ + 2 = 0, respectively), and solve the resulting quadratic formulas, you get the four complex solutions.
x⁶ = 2²
x = ±³√2, ±³√2(1+i*√3)/2, ±³√2(1-i*√3)/2
As an entrance exam, assuming the takers have only the minimally required highschool algebra, this is a very doable but long problem.. anybody with more advanced math experience will be able to do this in their sleep.
Repeating Over and Over 'cubes root of two...' is driving me crazy!
Hören lernen heißt auch weghören lernen.
x⁶-4=0
(x³)²-2²=0
(x³+2)(x³-2)=0
x³+2=0
x³=-2
x=³√-2 ❤
x³-2=0
x³=2
x=³√2 ❤
Four complex roots.
Which are?
But wouldn't cube root of -2 be undefined? Just wondering
@@Knak-z1y No. it’s negative cube root of 2.
But it's written cube root of negative 2
@@Knak-z1y ³√−2 = −³√2.
Interesting, but long winded. Why not just use De Moivre's theorem? Also, the question as posed does not state that the solution has to be real, but many people would take that as read. In that case the only extra points would be for spotting the negative solution. The question is not precise enough.
Because using rote learned algorithms is not actually conceptualization which is what an entrance exam is ultimately testing. Or ability to derive it on your own.
I think i'll go back watching Terrrence Howard bs videos
It’s a straight forward question 😅
You should go through IIT entrance exam, it has way more quality problems than this one.
Das ist immer richtig. Spiele sind aber Voraussetzung.
Problems like this required advanced techniques before numerical computing came along 50-60 years ago. In 15 sec, I get x= 1.259921049894873 from a scientific calculator. Some will call into question why Harvard still holds onto pre-computational challenges like this? Tradition?
Actually, this problem has been simple much longer, since about year 1750. Nothing more advanced than Euler's Formula required. In my head, I knew the answer was cube-root 2 times each of the 6 complex roots of unity.
@@KennethDuda OK, but if I need to solve this for a practical application, I still need numerics for the cube root of 2. I'm going to consult a calculator. So why not just enter the 6th root of 4? I still don't see the point of Euler's formula if I can calculate y√x.
@@pbierre If working on an engineering problem, especially if it's to do with vibrations/oscillations, one might need the complex roots as well. Also in acoustics, signal processing, classical waves, quantum mechanics they can be important...They might correspond to different phases of oscillations, different spacings of elements in a structure, different times an aftershock might come after an earthquake, different particles or particle states, myriads of things. But hopefully by the time anyone started working on any of that stuff they'd know all about the polar form of complex numbers and Euler's formula and could go straight to the answer instead of doing all that tedious algebra.
@@pbierre Yep, what zetacrucis681 said --- the problem has six solutions, and they might all have practical relevance depending on why we're trying to solve x^6 - 4 = 0. So we might need all six.
Typically, in mathematics, "solving" something means reducing it to a form "x = " where the expression involves well-known constants and functions. Thus, x^6 - 4 = 0 is not solved, but x = (2^(1/3)) * e^(k * pi * i / 3) for k = 0,1,2,3,4,5 would be considered solved. Thus you don't "need numerics" to get to the solution of this problem. If you want a numerical approximation for x, then we can get out our calculators etc., but it might be worth keeping in mind that once you enter 4^(1/6) in your calculator, you actually no longer have the exact right answer because there is no precise (finite) decimal representation of that number.
What is the point of all this? That's a question for philosophers, not mathematicians!
Why is that difficult ??
I got answer without solving, just calculate x= 1.26
But there are really six roots as he said. I like using Euler’s method the best.
Boy, he sure made a meal out of that. If X6-4=0, then it seems to me that x = the cube root of 2. No ? Why not ?
There are five other solutions. Take the 6 complex roots of unity, and multiple each by the cube-root of 2, and those are the six solutions. (Note that -1 and 1 are two of the six complex roots of unity, yielding the real solutions 2^(1/3) and -2^(1/3), but there are four other roots of unity, yielding four more solutions. One of those is x = 2^(1/3) * (cos pi/3 + i * sin pi/3).)
At 7:15, you should have used a different variable from x since x was already the variable in the original equation! Also, I do not like mixing radical and exponential forms in the same expression.
2 ^ (1/3) is one answer.
Can you Solve the Equation 3x+5=3(x+2) Fast!?
😊أول مرة أرى معادلة من درجة ثانية يوجد فيها مجهولينx,x^2 تحل بديلطا
"Higher Mathematics", this is totally obvious, it's just each of the 6 complex roots of unity, times the cube-root of 2. I could do that in my head. Why the long video? Why difference of squares and sum of cubes formula? None of this is needed. You are making an easy problem hard.
👍
If x⁶, then this problem have 6 solutions, so we need to find all of them, not just 1
Sie benutzen vorgefertigte Ergebnisse; das ist nicht der Sinn der Sache. Herleiten ist die Aufgabe.
@@daangueto 6 komplexe Wurzeln sind alle. Mit realen Sonderfällen.
@@daangueto Yes, the six solutions are 2^(1/3) * e^(k * pi * i / 3) for k = 0, 1, 2, 3, 4, 5. Sorry, that was what I meant by "each of the 6 complex roots of unity, times the cube-root of 2".
x^6 = (4^(1/6))^6.
x = 1.26
This can be done by observation. Becomes x^3 = 2 and x^3 = -2. Then the 3
roots of these, which are trivial.
And only 2/6 of the answer.
(x^6)^2 ➖ (4)^2= {x^36 ➖ 16}=x^20 x^2^10 x^2^2^5 x^1^2^1 x^2^1 (x ➖ 2x+1).
Good
@@Grammarforeveryone thank you
Why not solving it using polar coordinates?
x⁶ - 4 = 0
x⁶ = 4
Imagine x is a complex number, such as: x = a + ib.
Suppose that: x⁶ = 6
The module of x⁶ is: M = 4 → the module of x ix: m = M^(1/6)
m = 4^(1/6)
m = [2^(2)]^(1/6)
m = 2^(2/6)
m = 2^(1/3) or m = ³√2 ← this is the module of x
The argument of x⁶ is β such as: tan(β) = 0 → β = 0 → similar to 2π
The argument of the first root of x is: β/6 = 2π/6 = π/3
So we can see that the first root of x⁶ is:
x1 = m * [cos(π/3) + i.sin(π/3)] → to get the second root, we add (2π/6) to the initial angle, i.e. (π/3)
x2 = m * [cos(2π/3) + i.sin(2π/3)] → and so on to the following
x3 = m * [cos(3π/3) + i.sin(3π/3)] → …
x4 = m * [cos(4π/3) + i.sin(4π/3)] → …
x5 = m * [cos(5π/3) + i.sin(5π/3)] → …
x6 = m * [cos(6π/3) + i.sin(5π/3)]
Recall: m = ³√2
x1 = 2^(1/3) * [cos(π/3) + i.sin(π/3)]
x1 = 2^(1/3) * [(1/2) + i.(√3)/2]
x1 = 2^(1/3) * (1/2) * (1 + i√3)
x1 = 2^(1/3) * 2^(- 1) * (1 + i√3)
x1 = 2^(- 2/3) * (1 + i√3)
x2 = 2^(1/3) * [cos(2π/3) + i.sin(2π/3)]
x2 = 2^(1/3) * [(- 1/2) + i.(√3)/2]
x2 = 2^(1/3) * (1/2) * (- 1 + i√3)
x2 = 2^(- 2/3) * (- 1 + i√3)
x3 = 2^(1/3) * [cos(π) + i.sin(π)]
x3 = 2^(1/3) * [(- 1) + i.(0)]
x3 = - 2^(1/3)
x4 = 2^(1/3) * [cos(4π/3) + i.sin(4π/3)]
x4 = 2^(1/3) * [(- 1/2) + i.(- √3)/2]
x4 = 2^(1/3) * (1/2) * (- 1 - i√3)
x4 = - 2^(- 2/3) * (1 + i√3)
x5 = 2^(1/3) * [cos(5π/3) + i.sin(5π/3)]
x5 = 2^(1/3) * [(1/2) + i.(- √3)/2]
x5 = 2^(1/3) * (1/2) * (1 - i√3)
x5 = 2^(- 2/3) * (1 - i√3)
x6 = 2^(1/3) * [cos(2π) + i.sin(2π)]
x6 = 2^(1/3) * [(1) + i.(0)]
x6 = 2^(1/3)
To go further:
= 2^(- 2/3)
= 2^[2 * (- 1/3)]
= [2^(2)]^(- 1/3)
= 4^(- 1/3)
= 1/4^(1/3)
It's an easy problem done in just 16 sec
Also kein Problem.
On y arrive aussi en recherchant les n racines d’un nombre complexe z:
z^n=r.e^i.2kpi/n avec k E {0;n-1} avec r=4 et n=6
Ce qui donne:
z^6=4.e^i 2kpi/6 avec k E {0,5}
Cela oblige à repasser à la forme a+ib:
z=r sin téta + i r cos téta
Avec r=4^1/6 et téta=pi/3 et ses multiples.
Il est vrai que ça tombe bien car, dans ce cas, les 2kpi/6 tombent sur des valeurs dont on connais les sin et cos.
J’accepte volontiers les critiques 😢
X^6 - 4 = 0
X^6 = 4
X = 6√4 ( 6 here is used to show the order of the surd)
X = 6√2^2
X= 3√2 ( again ,3 here is the order of the surd)
No need of all that complicated stuff
love uuuu
You are missing 5 out of 6 answers.
@@Nikioko true , I guess. An equation of the degree 6 must have 6 solutions .
@@horrifichalo8430 why should there be 6 solutions if any positive integer exponentiation multiple of two is a parabolic function and has only 2 points of intersection with a linear function?
@@quadrat_1 i have no clue what those are ...
only thing i understand is that the graph is a parabola cause of the even power
So Easy (
Warum nicht den Satz vom Nullprodukt anwenden?
So würde ich lösen:
x⁶ − 4 = 0
3. binomische Formal anwenden:
(x³ − 2) ⋅ (x³ + 2) = 0
Satz vom Nullprodukt ergibt schon mal:
x₁ = ³√2 ∨ x₂ = −³√2
Diese Lösungen kann man jeweils aus den kubischen Funktionen ausklammern:
(x − ³√2) ⋅ (x² + ³√2 x + ³√4) ⋅ (x + ³√2) ⋅ (x² − ³√2 x + ³√4) = 0
Und nun kann man die beiden quadratischen Funktionen (wieder Satz vom Nullprodukt) mit der pq-Formel lösen, um die vier komplexen Lösungen zu erhalten:
x₃,₄ = −³√2 / 2 ± √(³√4 / 4 − ³√4)
= −³√2 / 2 ± ³√2 ⋅ i ⋅ √3 / 2
= −³√2 / 2 ± ⁶√108 ⋅ i / 2
x₃ = (−³√2 − ⁶√108 ⋅ i) / 2 ∨ x₄ = (−³√2 + ⁶√108 ⋅ i) / 2
x₅,₆ = ³√2 / 2 ± √(³√4 / 4 − ³√4)
= ³√2 / 2 ± ³√2 ⋅ i ⋅ √3 / 2
= ³√2 / 2 ± ⁶√108 ⋅ i / 2
x₅ = (³√2 − ⁶√108 ⋅ i) / 2 ∨ x₆ = (³√2 + ⁶√108 ⋅ i) / 2
Dann kann man noch die Lösungsmenge mit den sechs Lösungen angeben, aber das schenke ich mir hier mal. Genau wie die Probe.
Well, I got cube root of two without pencil and paper.
X^6-4=0
6 log(x) = log(4)
Use base 2 logs.
6 log(x)=2
Log(x)=1/3
X= 2^1/3. Plus or minus since we’re raising x to an even power in the original equation.
Which is just one third of the answer.
@@Nikioko please share. I found plus and minus cube root of 2. Is there another number that equals 4 when raised to the 6th power? I would like to learn. The square root of minus one mystifies me.
@@johnnyragadoo2414 x⁶ − 4 = 0
Third binomial formula a² − b² = (a + b) ⋅ (a − b):
(x³ − 2) ⋅ (x³ + 2) = 0
Rule of the zero product: The product is zero when at least one of the factors is zero, so:
x³ − 2 = 0 or x³ + 2 = 0
And thus:
x³ = 2 or x³ = −2
By taking the third root, we get:
x₁ = ³√2 or x₂ = −³√2
Let's take the first factor of (x³ − 2) ⋅ (x³ + 2) = 0:
x³ − 2 = 0 for x = ³√2, obviously.
So let's factor out x − ³√2:
(x − ³√2) ⋅ (x² + ³√2 x + ³√4) = 0
Again, rule of the zero product. We know that with x = ³√2, the first factor becomes zero. So let's make the second factor zero:
x² + ³√2 x + ³√4 = 0
Now you can solve the quadratic formula the way you prefer:
x₃,₄ = −³√2 / 2 ± √(³√4 / 4 − ³√4)
= −³√2 / 2 ± √(−3 ⋅ ³√4 / 4)
= −³√2 / 2 ± √−3 ⋅ √(³√4 / 4)
= −³√2 / 2 ± i ⋅ √3 ⋅ (³√2 / 2)
= ³√2 / 2 ⋅ (−1 ± i ⋅ √3)
Now you have two complex solutions.
If you factor out x + ³√2 on x³ − 2 = 0 and solve the quadratic formula again, you get the other two complex solutions:
x₅,₆ = ³√2 / 2 ⋅ (1 ± i ⋅ √3)
So, overall, you get the two real solutions x₁,₂ = ±³√2 and the four complex solutions x₃,₄,₅,₆ = ³√2 / 2 ⋅ (±1 ± i ⋅ √3).
@@Nikioko Yep, I gotta figure out i. Thank you!
x^6=4
6*ln(x)=ln(4)
ln(x)=ln(4)/6. e^
x=e^ln(4)/6=1,259
1,259^6=4
8:59
Can anyone explain how 2^(2/3)-4*2^(2/3) equals -3*2^(2/3)
For any x, x - 4x = -3x. This works when x=2^(2/3).
10^1.1= 1.26 10^1.2=1.58 10^1.3=1.98 10^1.4=2.51 10^1.5=3.16 10^1.6=4. That's it. 1.26 x 6 equals 4
Jhama Likhlam
Tui
T to begin with
Point 1 to 2 going 90deg covering 0deg there wee go a b c... And so many languages
Now let's lang one let's g
Get it
Uff
1.2599
Si a et b sont des nbres complexes, avec, en polaires,
a =(|a|,arg(a)) et arg(a) € [0;2pi[, et b = x+i.y, on a, en polaires,
a^^b = ( exp(x.ln(r) - y.arg(a)) , x.arg(a) + y.ln(r) ), ceci est vrai ¥a et ¥b (pourvu que arg(a) € [0, 2pi[).
Ici a^^6 = ( exp(6.ln(|a|) , 6.arg(a) ) et 4 = ( 4 , 0 ) (tout en polaires) ==> exp(6.ln(r)) = 4 et 6.arg(a) = 0 + 2k.pi.
D'où les 6 racines.
Rq ln est le logaritme réel.
X^6=4
X= 6th root of 4
Усложнено донельзя!
x^6 - 4 = (x^3 + 2)(x^3 - 2) = (x + (2)^(1/3))(x^2 - (2)^(1/3)x + (2)^(2/3))(x - 2^(1/3))(x^2 + (2)^(1/3)x + (2)^(2/3)) = 0
x = +/- (2)^(1/3), ((2)^(1/3) +/- ✓((2)^(2/3) - (2)^(7/3)))/2,
(-(2)^(1/3) +/- ✓((2)^(2/3) - (2)^(7/3)))/2
So x=21/3
So y=😅
Now z=😢
What?????
Looks weird. Cause no one needs to take the answer with complex numbers. And every time all solutions are based on a simple rule - "D" does not equal "- value". So, X 3456 no need here. In the real world it calls - over engineering
Es gibt keine Welt, die wir real verstehen. Unterkomplexes Denken.
3s bố có kết quả rồi mà thằng viện sĩ toán học vẫn còn đang giải
Stop at real roots. Complex roots equals no roots.
Why should one not include complex roots?
Because@@demoman1596sh