Wow! Thanks for sharing this wonderful appreciate I will study it carefully and closely at my leisure time so that I can apply it in solving similar challenges of this kind in subsequent video tutorials sir.
It is usually required to solve equations in real numbers, unless the condition specifically states that complex roots are needed. It is not clear why p,q = 1, -3, which can give real roots, are rejected.
If a is isolated in the left side And b and numbers in rigth side (8/4)^a=(..)/(..) 2^a=(..)/(..)will be introduced in first equation. This was in '75. French algebra translated books were simple
Hahahaha....anyway thanks for dropping this comment sir. You made my day as I read you comment and laughing at this same time sir. Much love to you sir... 💕💕💖💖🥂🥂
Brother, you have done so well but there's a little mistake that needs to be corrected, even though it didn't affect the answer. It has to do with the factorization of the quadratic equation p²+p -20=0. The parentheses in the next step are wrongly placed. That's (p²+5p) - (4p -20) =0 is supposed to be written as (p²+5p) + ( -4p -20) =0 Remember that if we expand these two equations we won't get the same result
Thanks for appreciating our works here sir and thanks for the correction, it is well noted and I will be more careful in subsequent lectures sir. Maximum respect sir...👍👍💪💪💪
Sure sir, we all here are having a blissful morning. We hope you are having same over there, we wish you all the best in this season also sir. Much love from all of us at Onlinemathstv to you sir. 💕💕❤️❤️💖
No sir, the minus at the back of 4 will change the plus sign to minus, so that expression is 100% correct. Thanks for watching and dropping this comment. Much love ❤️❤️💕💕
Really? we will appreciate it more if and only if you can point them out steps by steps so that we can amend and avoid such errors in the new video tutorial sir. Thanking in you in anticipation.
Sir you are true genius, i was expecting that this problem cannot be solve but you solve this with ease
Support from India 🇮🇳❤
Gracias estuvo muy bueno, hubo de todo; me hubiese gustado que no terminará en un número imaginario. Saludos desde México 👏👏👏Merry christmas.
Bravo! Nice work. So when all is said and done there are no solutions in the real world because of the iota.
No sir.
Hi and thank you for your video.
Another useful approach after minute 10:41 coud be:
EQ1: (x+y)^3-3xy(x+y) = x^3 + y^3 = 10
EQ2: (x+y)^2-2xy = x^2 + y^2 = 7
-------------------------------
Set EQ1 = EQ2 by xy
[(x+y)^2 -7]/2 = xy = [(x+y)^3-10]/[3(x+y)]
-------------------------------
substitute (x+y) = u
-------------------------------
3u(u^2-7) = 2(u^3-10)
-------------------------------
u^3-21u+20
find first solution and after that do division and solve qudratic equation
u1=1; u2=4, u3=-5
solve u: 6 solutions
-------------------------------
=>
x^2+y^2 = 7
from u=4:
x+y = 4
y = 4-x
x^2 + (4-x)^2 = 7
2x^2 - 8x +9 = 0
x12 = [8±j*sqrt(8)]/4
resubstitute
Wow! Thanks for sharing this wonderful appreciate I will study it carefully and closely at my leisure time so that I can apply it in solving similar challenges of this kind in subsequent video tutorials sir.
Entendido. Gracias por el video. Saludos desde Chiclayo Norte del Perú
we are glad you understood and gained some values from this, Thanks for watching and dropping this comment sir.
Əla həll etdiniz.Bakıdan salamlar.
wow! Thanks a bunch for this acknowledgement sir.
Much love from all of us here at Onlinemathstv to you sir...❤️❤️💕💕😍😍
Lovely stuff...
This is a hangover BURSTER!
Thanks a million for watching and dropping this encouraging and wonderful comment here sir.
Much love deeply💕💕
Profeseur is briliant student teaching people
hahahaha...my good friend
Thank you. Suppose very clever at maths
You are most welcome sir and thanks for this encouraging comment.
We love you so so much sir...❤️❤️❤️
I like this problem. Good explanation
...... I think I need a drink. That was epic. Thank you.
Oh, by the way, Merry Christmas!😅. You are a great teacher.
Thank you! 😃 We all at Onlinemathstv you the very best in this festive season sir.
This is mathematics. Thanks a big deal master.
This is beyond my scope I beg, anyway thanks for the video I enjoy your explanations teacher. Wishing the best of the season....🎉🎉🎉❤❤❤
Hahahahaaha....that is not true my good friend.
Just give it a detailed study for few times and you will discover it is easier than you thank sir/ma.
I found all 6 complex solutions, it took me about 1 hour to sovle all
Bravo!!! You are the master here,
maximum respect to you sir.
It is usually required to solve equations in real numbers, unless the condition specifically states that complex roots are needed. It is not clear why p,q = 1, -3, which can give real roots, are rejected.
Excellently explained. But we can minimize writing time by not repeating, thus shorten the video. Thanks for uploading. 2024.3.26
Wonderful
Thanks a million sir.
If a is isolated in the left side
And b and numbers in rigth side
(8/4)^a=(..)/(..)
2^a=(..)/(..)will be introduced in first equation.
This was in '75.
French algebra translated books were simple
Noted, thanks a million sir.
Good work.Very nice problem, very ugly solution. Exactly the opposite of what Ramanujan would have done. Or not.😊
Hahahaha....anyway thanks for dropping this comment sir.
You made my day as I read you comment and laughing at this same time sir.
Much love to you sir... 💕💕💖💖🥂🥂
Brother, you have done so well but there's a little mistake that needs to be corrected, even though it didn't affect the answer. It has to do with the factorization of the quadratic equation
p²+p -20=0. The parentheses in the next step are wrongly placed.
That's (p²+5p) - (4p -20) =0 is supposed to be written as
(p²+5p) + ( -4p -20) =0
Remember that if we expand these two equations we won't get the same result
Thanks for appreciating our works here sir and thanks for the correction, it is well noted and I will be more careful in subsequent lectures sir.
Maximum respect sir...👍👍💪💪💪
굿굿
Thanks for appreciating our little effort thus far.
I didn't understand why you rejected (p,q) with negative p or negative q...
If you admit complex solutions you cannot reject (p1,q1) and (p3,q3)
Hello hope you are having a good morning 🌄 merry Christmas Eve
Sure sir, we all here are having a blissful morning. We hope you are having same over there, we wish you all the best in this season also sir. Much love from all of us at Onlinemathstv to you sir. 💕💕❤️❤️💖
Horner's algorithm seems to me to be better than polynomial division. Less writing, and a faster way to the zeros of a polynomial.
Ok, I will apply that in subsequent challenges sir/ma.
1^1 2^5 (x+1x-1) (x+2x-5)
Wish you merry Christmas 🎅 🎄
Many thanks for the wishes and all of us at Onlinemaths tv love you dearly for this comment sir.
Marry Christmas ❤🎉
Wishing you and your entire family the best of the season sir.
19:07 - If you multiply 4*(p+5) it will be equal to 4p+20 not 4p-20.
No sir, the minus at the back of 4 will change the plus sign to minus, so that expression is 100% correct. Thanks for watching and dropping this comment.
Much love ❤️❤️💕💕
This is complexes math
Hahahaha...🤣🤣😂😂 Yes it is complex but can be learnt in few steps, just follow the steps carefully and you are good to go.
Thanks sir.
last line is irror
So so sorry about that, I will be more careful in subsequent time master. Thanks for the good observation sir.
Hi
WHO and what is mathematics ???
🌈🌈🌈🌈🌈
too many errors. Redo.
Really? we will appreciate it more if and only if you can point them out steps by steps so that we can amend and avoid such errors in the new video tutorial sir.
Thanking in you in anticipation.
You are wicked😂