you can raise to the power of 6 it will cancel the cube root and the square root giving you after simplifying the sum of two positive terms being equal to 0 which means they're both 0 hence x=0
If you take the 6th power of both sides, you have a quadric equation, 3z4-2z3+3z2=0 with 0 as double solution and 2 complex conjugate solutions. However by definition of the complex roots (real part always positive), one can check that the two complex solutions are not solutions of the original equation. By the way ... a = b^(1/2) and a^2 = b are not always equivalent ...and also a = b^(1/3) and a^3 = b For example ... (-1)^3 = -1 ... everybody knows ... but if you calculate (-1)^(1/3) in a complex calculator, you obtain ... 0.5 + 0.866j
From a first glance, it's obvious x=0 is the only real solution. However, for the complex solutions I got x=((-7+-2*i*sqrt(2))/9)^3. Does that look correct?
просто возведи обе части в шестую степень, сократи подобные члены и сделай замену корень шестой степени из x =y. Вновь сократи и получи квадратное уравнение без действительных корней. Зато найдешь мнимые.
Exactly ! That's exactly what I did ... and by elevating to the power of 6 you obtain a quadric with 0 as double solution and 2 complex conjugate. However you have to check the solutions to the original equation, and so you should reject the non-0 solutions that are not solutions of the original equation.
I finally got the 1 year badge!!! I love your videos, and I hope to get to 2 years!! Keep making AMAZING videos!
Wow!!! Congratulations and thank you for the support! 🤩💖
How did yo get the badge?
you can raise to the power of 6 it will cancel the cube root and the square root giving you after simplifying the sum of two positive terms being equal to 0 which means they're both 0 hence x=0
wow i like these problems that includes radicals
Only to congratulate you for your math channel.
I love many of your problems.
Thank you! 💕
you always surprize even though i watched this video one time
73 / 5.000
what hardware and software do you use to write on this digital blackboard ?
It's the Perry the Platypus colour.
A visual example, as of a simple thing to make a difficult, and nothing to learn, but firmware
Before to watch the video: My first method was guessing ('simple inspection') xd Second method: x=z^6; solve for z {idem than yours :) }
Do the compex solutions from that equation work?
There are no other solutions than x=0, complex or not.
@@souvik9632 No, there are NOT two solutions. The only solution is x = 0.
@@souvik9632 There is only one solution to the given equation, x=0. There are no other solutions, real or complex.
If you take the 6th power of both sides, you have a quadric equation, 3z4-2z3+3z2=0 with 0 as double solution and 2 complex conjugate solutions.
However by definition of the complex roots (real part always positive), one can check that the two complex solutions are not solutions of the original equation.
By the way ... a = b^(1/2) and a^2 = b are not always equivalent ...and also a = b^(1/3) and a^3 = b
For example ... (-1)^3 = -1 ... everybody knows ... but if you calculate (-1)^(1/3) in a complex calculator, you obtain ... 0.5 + 0.866j
I modified the equations a bit… and graphed y=(x^3-1)^2 and y=(x^2-1)^3, kinda cute… confirming the same intersection btw…
Perfect solving 👌
Thank you 💕
Thanks!!
You’re welcome!
Slight mistake at the end: the intersection is actually at (0,1).
Yeah
But that's right: for x = 0, y=1
It is mildly interesting that I enjoy your PJ's (don't ask 'y', '2b' or not '2b', '2u', and so on...) 🤷♂
Not me finding complex solutions and using moivres formula 😭😭
From a first glance, it's obvious x=0 is the only real solution. However, for the complex solutions I got x=((-7+-2*i*sqrt(2))/9)^3. Does that look correct?
No, x=0 is the only solution, complex or otherwise.
❤❤❤❤
просто возведи обе части в шестую степень, сократи подобные члены и сделай замену корень шестой степени из x =y. Вновь сократи и получи квадратное уравнение без действительных корней. Зато найдешь мнимые.
Верно!
i got some random strange answers
which is x = (2114±√2114^2-4) /2
You have a Turkish accent, are you Turkish by any chance ?
Yes he is turkish
Both methods are too sluggish. Simply raise both sides to the power of 6 and see how it goes.
Exactly ! That's exactly what I did ... and by elevating to the power of 6 you obtain a quadric with 0 as double solution and 2 complex conjugate. However you have to check the solutions to the original equation, and so you should reject the non-0 solutions that are not solutions of the original equation.
@@tontonbeber4555 "quadric?" You mean "quadratic," correct?
@@forcelifeforce I think he meant quartic.