Thank you for your time, you are really doing good tutorials. I cannot spot my mistake, if any. Any comment will be welcome. Let us consider X2 and X3 both in polar form... 3(45⁰) means modulus=3 and argument=45⁰ X2 = 1/2 (-1 + i sqrt(3) --> 1(120⁰) ( I'm not writing the usual ±k•pi) X3 = 1/2 (-1 - i sqrt(3) --> 1 (-120⁰) Plugging these values in X • sqrt(X) + 1 = 0 we find For X2: 1(120⁰) • sqrt [ 1(120⁰) ] + 1 = 0 1(120⁰) • 1(60⁰) + 1 = 0 1(180⁰) + 1 =0 - 1 + 1 ≠ 0 For X3: 1(-120⁰) • sqrt [ 1(120⁰) ] + 1 = 0 1(-120⁰ ) • 1(-60⁰) + 1 = 0 1(-180⁰) + 1 =0 -1 + 1 ≠ 0 So both X2 and X3 are solutions. It is logical because when a polynomial with real coefficients has a complex root, its conjugate is also a root.
Thank you sir for your very interesting videos. I am an electric engineer who has always been very passionate about mathematics. I really like the simplicity and rigor with which you explain the concepts. I also subscribed to your channel because my son who is studying at scientific high school appreciates your work very much. Warm greetings from Italy!👍👍👍👍💪💪💪👋👋👋
You are most welcome sir and thank you for subscribing to our channel. For the sake of people like you and others who believe in what we do here we will try our possible best ability to give the best in order not to disappoint you and others who believe in us sir. Once again, thanks and we all @onlinemathstv love you dearly sir. ❤️❤️❤️💖💖💖💕💕💕
Thanks for the compliment sir. As for your question, I have solved this math challenge previously, you can watch the video tutorial that actually produce the x1, x2 and x3 via the link below sir. ua-cam.com/video/VEnRUz2T8c4/v-deo.html It is well detailed there. Thanks and much love ❤️❤️❤️💖💖💕💕
Nice 👍
thanks boss
Nice video 😊 I AM from India ,TN you rocking man 🎉
What grade his teaching now?
thanks sir.
Thank you for your time, you are really doing good tutorials.
I cannot spot my mistake, if any. Any comment will be welcome.
Let us consider X2 and X3 both in polar form... 3(45⁰) means modulus=3 and argument=45⁰
X2 = 1/2 (-1 + i sqrt(3) --> 1(120⁰) ( I'm not writing the usual ±k•pi)
X3 = 1/2 (-1 - i sqrt(3) --> 1 (-120⁰)
Plugging these values in X • sqrt(X) + 1 = 0 we find
For X2:
1(120⁰) • sqrt [ 1(120⁰) ] + 1 = 0
1(120⁰) • 1(60⁰) + 1 = 0
1(180⁰) + 1 =0
- 1 + 1 ≠ 0
For X3:
1(-120⁰) • sqrt [ 1(120⁰) ] + 1 = 0
1(-120⁰ ) • 1(-60⁰) + 1 = 0
1(-180⁰) + 1 =0
-1 + 1 ≠ 0
So both X2 and X3 are solutions. It is logical because when a polynomial with real coefficients has a complex root, its conjugate is also a root.
Thank you, the process seems simple at first, but it really is a hell of an equation
Smiles...Thanks for watching sir.
Much love 💕💕💕💕
Thank you sir for your very interesting videos. I am an electric engineer who has always been very passionate about mathematics. I really like the simplicity and rigor with which you explain the concepts. I also subscribed to your channel because my son who is studying at scientific high school appreciates your work very much. Warm greetings from Italy!👍👍👍👍💪💪💪👋👋👋
You are most welcome sir and thank you for subscribing to our channel. For the sake of people like you and others who believe in what we do here we will try our possible best ability to give the best in order not to disappoint you and others who believe in us sir.
Once again, thanks and we all @onlinemathstv love you dearly sir. ❤️❤️❤️💖💖💖💕💕💕
Thanks for making this video sir, because am one of those confused with the simplification after substitution of the the two imaginary roots.
You are welcome sir
I've enjoyed following you're teaching.
Question- how did you originally come up with the X1, x2 & x3 values? Thank you.
Thanks for the compliment sir. As for your question, I have solved this math challenge previously, you can watch the video tutorial that actually produce the x1, x2 and x3 via the link below sir.
ua-cam.com/video/VEnRUz2T8c4/v-deo.html
It is well detailed there.
Thanks and much love ❤️❤️❤️💖💖💕💕
.x^1,5=-1; x=(-1)^2/3=1^(1/3)
Only x = 1 doesn't satisfy the equation but the rest do sir
Why didn't you take √1=-1? In this case, 1 also satisfies the original equation. 1*-1+1=0
We need to simplify the L. H. S and compare with the R .H. S
x√x +1 = L. H. S kindly simplify this first and later compare it with the R H. S = 0
I got brain damage with your problem sir😂
Explicit
Thanks sir