Complex Numbers : Roots of a cubic equation : ExamSolutions
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- Опубліковано 12 лют 2012
- Complex numbers: finding the roots of a cubic equation.
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11 years on and this video is a blessing- Thank you!
Exactly what I needed! Thank you so much for the perfect explanation
Thank you, that has greatly helped my understanding
thank you so much for the upload sir! i wasn't aware of such a straightforward method!
Thank u so mch sir..this was exactly what I was looking for..
Very nicely taught, thank you very much Sir!
if only my maths teacher had described it like this.so much easier to understand now thank you.
Thank you so much!
wow superb!!its applicable to all numerical. i have been searching since many days and i got it finally..thanks lot u sir.
+Parshu Ram Well done.
That was very help sir. Thank you.
This is a great video. Keep it up!
Thank You So Much :) This really helped me a very lot ! Thanks Again !
Thilak Kumara Thanks for using.
Thanks soooo much!
You're welcome. Thanks for using my site.
ExamSolutions
Another fantastic video sir.
Can I just ask you is the only way to get the quadratic factor by long division, I used the method but was wondering is there an alternative method.
And also how would you obtain the roots of an equation such as x^4-3x^3+2x^2-3x+2?
Regards sir :)
@@ExamSolutions_Maths YOU ARE THE BEST!!
a trick : you can watch movies on Flixzone. Been using them for watching a lot of movies these days.
@Van Brecken yea, I have been watching on Flixzone for months myself :D
Hi sir, this has been very helpful. Do you mind doing more IB videos? It will be truly appreciated!
best video on this topic really useful
For the roots to be a complex conjugate pair the equation must consist of terms of the form az^3+bZ^2+cZ+d=0 where a does not equal zero and a, b, c, d and are real. d for this equation is imaginary and so this rule does not hold. Your roots are correct though.
Thx was very helpful
thankyou so much!!
No problem.
at 4:20 you say f(1)=22 but I got 24 on my calculator. And it is the same for -1
Thank yo Sir!
when you write down the quadratic formula, I didn't see there is no minus in front of b.
Is the formula like that? (-b±square root of b^2-4ac)/2a?
b is minus 6, so the negative that is always in the formula cancels this negative out to give positive 6. Hence why there is no negative sign when substituted in to the formula.
For f(1), the answer is 24 not 22. However, great video!! Thanks!
thx!
Very good
Use one of following substitutions to get quadratic equation
Z=U+V
Z=U+\frac{7}{3U}
You can also get formula for trigonometric function of triple angle
with substitution like this
Z=2\sqrt{-\frac{7}{3}}\cos{\theta}
Thank you so much, god bless you good sir.
Thank you and you too.
is there a way to this without trial and error
What should I do with the remainder if I get it in a question?
what if we need to find roots in terms of w ?
What if none of the factors equate the equation to 0
You're an absolute life saver, thankyou so much for these videos!
Saving lives feels good. Thank you and best wishes.
exams in 13hrs and i comepletely forgot about this topic thank you
Always welcome
@@ExamSolutions_Maths aced those complex numbers questions! Idk about the rest tho…
And in this case: x^4+2x^3-x^2+2x+1, how to know its roots?
indeed,......
❤️❤️❤️🙏🙏
There should be
Z=U-\frac{7}{3U} instead of
Z=U+\frac{7}{3U}
It appears as though I must guess
After plugging in numbers (Guessing) it worked.
I think I just found the Gordon Ramsay of math
*Quick Fact*
If a cubic polynomial given has 2 real roots, those two must be identical.