I finally got the 1 year badge!!! I love your videos, and I hope to get to 2 years!! Keep making AMAZING videos!
@SyberMath
2 жыл бұрын
Wow!!! Congratulations and thank you for the support! 🤩💖
@Drk950
2 жыл бұрын
How did yo get the badge?
@karimjemel7405
2 жыл бұрын
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
@SuperYoonHo
2 жыл бұрын
wow i like these problems that includes radicals
@hans-rudigerdrzimmermann
2 жыл бұрын
Only to congratulate you for your math channel. I love many of your problems.
@SyberMath
2 жыл бұрын
Thank you! 💕
@elmachevera
2 жыл бұрын
73 / 5.000 what hardware and software do you use to write on this digital blackboard ?
@hornet-e9g
2 жыл бұрын
you always surprize even though i watched this video one time
@Vitzyk
2 жыл бұрын
A visual example, as of a simple thing to make a difficult, and nothing to learn, but firmware
@GirishManjunathMusic
2 жыл бұрын
It's the Perry the Platypus colour.
@wesleysuen4140
2 жыл бұрын
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…
@Drk950
2 жыл бұрын
Before to watch the video: My first method was guessing ('simple inspection') xd Second method: x=z^6; solve for z {idem than yours :) }
@nameskizo_y7153
2 жыл бұрын
Perfect solving 👌
@SyberMath
2 жыл бұрын
Thank you 💕
@Alians0108
2 жыл бұрын
Do the compex solutions from that equation work?
@XJWill1
2 жыл бұрын
There are no other solutions than x=0, complex or not.
@XJWill1
2 жыл бұрын
@@souvik9632 No, there are NOT two solutions. The only solution is x = 0.
@XJWill1
2 жыл бұрын
@@souvik9632 There is only one solution to the given equation, x=0. There are no other solutions, real or complex.
@tontonbeber4555
2 жыл бұрын
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
@georgina4874
2 жыл бұрын
Thanks!!
@SyberMath
2 жыл бұрын
You’re welcome!
@InnocentNeuron
2 жыл бұрын
It is mildly interesting that I enjoy your PJ's (don't ask 'y', '2b' or not '2b', '2u', and so on...) 🤷♂
@Jalina69
2 жыл бұрын
Not me finding complex solutions and using moivres formula 😭😭
@oenrn
2 жыл бұрын
Slight mistake at the end: the intersection is actually at (0,1).
@abhinavbhutada9b484
2 жыл бұрын
Yeah
@robertocunha2857
2 жыл бұрын
But that's right: for x = 0, y=1
@tobymurs6495
2 жыл бұрын
You have a Turkish accent, are you Turkish by any chance ?
@metehan9185
2 жыл бұрын
Yes he is turkish
@NXT_LVL_DVL
Жыл бұрын
❤❤❤❤
@Vitzyk
2 жыл бұрын
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.
@SyberMath
2 жыл бұрын
Верно!
@scottleung9587
2 жыл бұрын
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?
@XJWill1
2 жыл бұрын
No, x=0 is the only solution, complex or otherwise.
@zhugzhuangzury
2 жыл бұрын
i got some random strange answers which is x = (2114±√2114^2-4) /2
@wesleydeng71
2 жыл бұрын
Both methods are too sluggish. Simply raise both sides to the power of 6 and see how it goes.
@tontonbeber4555
2 жыл бұрын
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.
@forcelifeforce
2 жыл бұрын
@@tontonbeber4555 "quadric?" You mean "quadratic," correct?
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