Thanks ... it reminds me of my studies ;) @16:50 we can use as well elegantly the complex roots of X^2+1 as follows ... F(X) = (10X^2-50X+10)/[(X-7)(X^2+1)] = A/(X-7) + (BX+C)/(X^2+1) Let X=7, then: F(X).(X-7) = A = (10x7x7-50x7+10)/(7x7+1) A = (10.49-50.7+10.1)/50 A = (50.10-50.7)/50 = (50.3)/50 A = 3 Let X=i first root of (X^2+1), then: F(X).(X^2+1) = Bi+C = (10i^2-50i+10)/(i-7) Bi+C = (-10-50i+10)/(i-7) = -50i/(i-7) = -50i(i+7)/(i^2-7^2) (I): Bi+C = -50(i^2+7i)/(-1-49) = -1+7i Let X=-i second root of (X^2+1), then: F(X).(X^2+1) = -Bi+C = [10(-i)^2-50(-i)+10]/(-i-7) -Bi+C = [-10+50i+10]/(-i-7) = -50i/(i+7) = -50i(i-7)/(i^2-7^2) (II): -Bi+C = -50(i^2-7i)/(-1-49) = -1-7i Using equations (I) and (II), as I+II, we get : 2C = -2 then C = -2/2 = -1 as I-II, we get : 2Bi = 14i then B = 14i/2i = 7 Finaly F(X) = (10X^2-50X+10)/[(X-7)(X^2+1)] = 3/(X-7) + (7X-1)/(X^2+1)
Am i even allowed to pick values for x to get rid of a,b or c ? Because technically for the second expression. 4 and 1 don't belong in the domain of definition
Yes, you are correct, thanks! I added an annotation to try to fix it, but those don't show up on all platforms. :( This will also make C = -2 and B = 1.
algebra error. -1+14-31=-18 therefore c=-2
Luis Aguirre Agreed. Thanks for spotting the error! I added an annotation.
yes he made a mistake dere
Damn i taught i wz wrong, thx dude :D
Thanks ... it reminds me of my studies ;)
@16:50 we can use as well elegantly the complex roots of X^2+1 as follows ...
F(X) = (10X^2-50X+10)/[(X-7)(X^2+1)] = A/(X-7) + (BX+C)/(X^2+1)
Let X=7, then:
F(X).(X-7) = A = (10x7x7-50x7+10)/(7x7+1)
A = (10.49-50.7+10.1)/50
A = (50.10-50.7)/50 = (50.3)/50
A = 3
Let X=i first root of (X^2+1), then:
F(X).(X^2+1) = Bi+C = (10i^2-50i+10)/(i-7)
Bi+C = (-10-50i+10)/(i-7) = -50i/(i-7) = -50i(i+7)/(i^2-7^2)
(I): Bi+C = -50(i^2+7i)/(-1-49) = -1+7i
Let X=-i second root of (X^2+1), then:
F(X).(X^2+1) = -Bi+C = [10(-i)^2-50(-i)+10]/(-i-7)
-Bi+C = [-10+50i+10]/(-i-7) = -50i/(i+7) = -50i(i-7)/(i^2-7^2)
(II): -Bi+C = -50(i^2-7i)/(-1-49) = -1-7i
Using equations (I) and (II),
as I+II, we get : 2C = -2 then C = -2/2 = -1
as I-II, we get : 2Bi = 14i then B = 14i/2i = 7
Finaly
F(X) = (10X^2-50X+10)/[(X-7)(X^2+1)] = 3/(X-7) + (7X-1)/(X^2+1)
The video was helpful and fascinating.
thanks
Michael.
Thanks very much for your detailed explanation.Was much needed.
thanks for providing tutorials for partials fractions.
Thank you!! This is the best video on partial fractions in a shorter time and effective way!!
thank you from new zealand i wasent taught this in class and it comes up frequently in tests. hats of too you
English next then XD
Thank you for the first at 0.04 slide which clearly states which problem appears at what time
Very useful to me thank you so much
13:43 👉 try pressing this in your calculator -1+14-31=-18 and not positive 😌😌
Exactly
What are you doing at min 14:38 what is that B(x^2-5x+4) where does it stem from?
That's from multiplying out B(x-4)(x-1)
@@MathWithMisterA yea i understood it after a while thanks a lot for the reply
can someone explain to me about 4:15? i dont really understand how to factoring that thing
I have some videos about factoring: ruclips.net/video/5gveQQq4ulY/видео.html
This is so helpful video thank you so much
MathwithmisterA good work i like the way you explain the maths. Thanx for that
thank you sir, very clear for me
THANK YOU SO MUCH ACUTELY YOUR EXPLANATION SO USEFUL
it's very helpful vedio for me so tks sir
Oh my god. U cleared up my day THANKS SOOOOO MUCH!
Am i even allowed to pick values for x to get rid of a,b or c ? Because technically for the second expression. 4 and 1 don't belong in the domain of definition
This is excellent, thank you so much!
thank you so much. very simple explanation :D
thank you so much good sir.
Thank you!
merci beaucoup!
Please tell me from which book you had took this question ....
thanks from India
thank you
thank you so much
Very helpful.. Thank u so much:)
Thankuuuuuu
wow.. It's of great help for me...
Thanks sir
it was awesome sir
You are the best 👏continue
How to solve this problem 1by x^4-1
13:41: -1^2 is 1, not -1
very useful.
thanks!!
Thankyou so much sir! your video has helped me a lot 👍👍
Is correct?
my answer: A=3 B=1 C =-2
For the 3rd example, yes, that is correct.
13:33 (-1+14-31)=-18
Yes, you are correct, thanks! I added an annotation to try to fix it, but those don't show up on all platforms. :(
This will also make C = -2 and B = 1.
Jiye bhutto
i love you
احسنت شرحك يخبل
Thanks for sharing! I posted a video on the cover-up method. Hope to get your thoughts.
Tqq sir
It's very helpful leaner s
Tq
-2 at 13:40
Yes, thanks! I put some annotations over that mistake a couple of years ago, but those annotations don't show up on all devices anymore. :(
no sound