Thank you sir I passed the exam of m2 with your valuable helpful video your method helped me too much thank you so much now I have a degree with 68 percentage thank you so much
Cardano became one of the most famous doctors in all of Europe, having treated the Pope. He was also an astrologer and an avid gambler, to which he wrote the Book on Games of Chance, which was the first serious treatise on the mathematics of probability
Here is another alternative approach. Let's factorise x³ - 27x + 54. x³ - 27x + 54 = x³ + (ax² - ax²) + (bx - bx) - 27x + 54 - (1) = (x³ - ax² + bx) + (ax² - (b + 27)x + 54) = x(x² - ax + b) + a(x² - (b + 27)x/a + 54/a) So, we'd like a² = (b + 27) and ab = 54 so that x² - ax + b = x² - (b + 27)x/a + 54/a and hence we can factorise the entire expression. So, we want b + 27 to be a square, and a to be a factor of 54. The first square larger than 27 is 36, so try b + 27 = 36, so b = 9 and a = 6. Looking good so far. So placing a = 6 and b = 9 in (1) then x³ - 27x + 54 = x³ + (6x² - 6x²) + (9x - 9x) - 27x + 54 = (x³ - 6x² + 9x) + (6x² - 36x + 54) = x(x² + 6x + 9) + 6(x² - 6x + 9) = (x + 6)(x² - 6x + 9) = (x + 6)(x - 3)² So the solutions are x = -6, 3 (double root).
It is rather a shrewd and ingenious way to fathom out roots of cubic equations. It could be solved by hit and trial, by contemplating one root and determining a factor which would then divide the equation to form a quadratic equation in form of quotient. I'm in class 10 and this seems very acquainted go me. Out of curiosity, who's the audience for the video?
Strictly it is an excellent approach to solve this type of quadratic equation. By putting x=3y this equation can be solved? I have divided it by x-1, x+1, x-2 and x+2 but there was a reminder. Can x+3 divide it?
An alternative approach is to used Descartes Rule of Sign and the Rational Root theorem. Descartes Rule of Signs says there are 0 to 2 +ve real roots and 1-ve real root. Possible rational roots by the Rational Root theorem are x = ±1, ±2, ±3, ±6, ±9, ±18, ±27 and ±54. I'd start checking the -ve roots, as at least we know one exist. When we try x = -6 the equation is satisfied and we can stop checking the negative roots. Then we just use synthetic division to find the quadratic factor, and then the quadratic formula to solve the quadratic equation.
Jab apan cardanos method lagate h... tab phle solution jo rhta h vo toh real rhta jabki baaki ke do bache hue solutions mh complex number bhi aata h naa .
Sir plzz solution of binomial equation kese karte hai or innke properties ka video banao jaldi se....exam ke liye jada din nhi hai so sir plzz bsc 1st sem solutions of binomial equation me vedio banao
Agr sir 2x^3 + x^2 + 1 ho qus tb isme missing term nhi hai pr x hai miss to isme me ese hoga solve jese apne kiya hai ki x^2 hi missing hona chahie ???
Hlo sir. I am Arun. Can you give the answer of my question. My question is that X3 +X2 + X - 1 = 0. The mean of X3 = X cube, X2= X square plus X minus 1 = 0 .Are u able to give the solution of my question at date 28/04/18 to 29/04/18.
If the value of (λ) is complex, Rational Numbers, Irrational Numbers How can it I solve ? Please make video Example (Q.1):x^4-4x^3+2x^2+12x=0 (Q.2):4x^4+2x^3-5x^2+6x-3=0 Please find roots by Ferrari's Method
Smooth explaination,and very helpful for me,. Thank you🌹
Wow sir ji bahut hi saral bhasa mai aapne kitni acchi taraha se is method ko samjha Diya bahut bahut dhanyavad ❤🔥🔥
Thank you sir I passed the exam of m2 with your valuable helpful video your method helped me too much thank you so much now I have a degree with 68 percentage thank you so much
m2?
sir bahut hi ache tarike se smjaya apne pehli bar me hi dimag mai baith gya
Cardano became one of the most famous doctors in all of Europe, having treated the Pope.
He was also an astrologer and an avid gambler, to which he wrote the Book on Games of Chance,
which was the first serious treatise on the mathematics of probability
Pucha kisine....
Did anyone asked you?
@@anujbhatt5933 just for likes
@Y337 being rejective to unnecessary information 😂🤣 u moron
Here is another alternative approach.
Let's factorise x³ - 27x + 54.
x³ - 27x + 54
= x³ + (ax² - ax²) + (bx - bx) - 27x + 54 - (1)
= (x³ - ax² + bx) + (ax² - (b + 27)x + 54)
= x(x² - ax + b) + a(x² - (b + 27)x/a + 54/a)
So, we'd like a² = (b + 27) and ab = 54 so that x² - ax + b = x² - (b + 27)x/a + 54/a and hence we can factorise the entire expression.
So, we want b + 27 to be a square, and a to be a factor of 54.
The first square larger than 27 is 36, so try
b + 27 = 36, so b = 9 and a = 6. Looking good so far.
So placing a = 6 and b = 9 in (1) then
x³ - 27x + 54
= x³ + (6x² - 6x²) + (9x - 9x) - 27x + 54
= (x³ - 6x² + 9x) + (6x² - 36x + 54)
= x(x² + 6x + 9) + 6(x² - 6x + 9)
= (x + 6)(x² - 6x + 9)
= (x + 6)(x - 3)²
So the solutions are x = -6, 3 (double root).
Kya yah important h
Looks complicated and lengthy, a good method, though
@@ambikanamboodiri8915
Thanks. I was just practising solving it a different way.
@@davidbrisbane7206 great 👍👍
Nice bro,,
👍
It is rather a shrewd and ingenious way to fathom out roots of cubic equations. It could be solved by hit and trial, by contemplating one root and determining a factor which would then divide the equation to form a quadratic equation in form of quotient.
I'm in class 10 and this seems very acquainted go me. Out of curiosity, who's the audience for the video?
This method is in B.tech and B.sc
@@civiluptodate1580 Oh, wow! It seems simple and easy nevertheless
Thanks sir I see ur previous videos of Ferrari n discarte methods.....i understand ....all rules 😊😊😊😊😊
Sir, what if we get a irrational root and cannot perform synthetic division?
Bhai mere khyal se (a+b)^3 ka formula a^3 + b^3 + 3(a^2)b + 3ab^2 hota hai.
Your writing is amazing. Alsi the method
Glad you think so!
😊😊😊
Sir your video are so interesting.
very nice explanation 🙏🏻
Keep watching
Very helpfull teaching aptitude.
👍👍👍👍👍👍very good job sir ji
What easy process to do this thnk u sir
Aapko easy laga😪😪
Strictly it is an excellent approach to solve this type of quadratic equation. By putting x=3y this equation can be solved? I have divided it by x-1, x+1, x-2 and x+2 but there was a reminder. Can x+3 divide it?
Yes.
video is very helpfull for me
Sir if there will be no perfect cube what to do then ?
Kaafi saral aur acchi explanation dii sir aapne thank you
Thanks sir❤ .This is examiner's favrt question.
Most welcome
Sir you are op please make more videos in maths
And also physics
Sir the Co efficient of x^3 should be 1 or if not then we have to take common the constant for comparison
This problem was too basic , easier with Rational root theorem , But i like this method 👍
Nice explanation 😀
Glad you liked it!
How come did a cube become a quadratic? Can't understand, can someone explain.
It s so easy with you thank you
Aree. Sir x equal u + v kyu let krte hai 😕
Sir what happened when the value of t is complex or negative iota?
Watch my another video, you will get your problem solved. Link :
ua-cam.com/video/BdIh3cDuC7c/v-deo.html
In this video, value of t is complex.
Thanks for explaining cardons method
Thank's. It's the only method to solve complex roots and decimal roots.
@Harshith Pandem hello
An alternative approach is to used Descartes Rule of Sign and the Rational Root theorem.
Descartes Rule of Signs says there are 0 to 2 +ve real roots and 1-ve real root.
Possible rational roots by the Rational Root theorem are x = ±1, ±2, ±3, ±6, ±9, ±18, ±27 and ±54.
I'd start checking the -ve roots, as at least we know one exist.
When we try x = -6 the equation is satisfied and we can stop checking the negative roots.
Then we just use synthetic division to find the quadratic factor, and then the quadratic formula to solve the quadratic equation.
thts too much effort , but good idea.
Bhai, apki videos wohot acchi hai.... pls Asymptotes karwado 🙏
Thank you sir for clearing the concept.
You are most welcome
Supbbbbbb concept sir ❤💙💙💙❤💙❤
You are great bro
Jab apan cardanos method lagate h... tab phle solution jo rhta h vo toh real rhta jabki baaki ke do bache hue solutions mh complex number bhi aata h naa
.
nice video, what if the complex roots of u and v are considered, then there are 9 solutions to x ?
It is a cubic equation, hence only 3 roots.
@@civiluptodate1580 you right, so what happens to the other values obtained ? would they be the same ?
Thank you sir for clear my concept
Great one
Sir cardons method to
B.sc maths honours me prhe the
I want Ferrari method how to do on this simple trick pls make a video on that topic
Videos on ferrari method are already present.
Check my channel.
Sir pta kaise chlega ki kitni baar synthetic division perform Krna h kyuki khi 3 bar toh khi 4 baar toh khi 1 baar kiya h jaise aapne abhi 1 baar kiya
From cubic equation to quadratic equation only one time.
Very helpful lesson but it was in English supposed to be more easier to understand but very helpful
Sir how tow take 't' in equation and 'x'
agar u aur v ki value complex ho to kaise solve karna hai???
Bhai achhe se samjhaya
Bahott achha sir 👍👍👍👍
cube root of -27 is not -3 .....
it must contain i term which is √-1
(-3)×(-3)×(-3)= -27
Same in reverse
It's very helpful for me🙂🙂
What if there would be -54 instead of +54. ?
Thanks sir ji !!
Awesome sir
Sir sum of zeroes to given hai hi ni wo to inke cube ka sum hai sir
how to solve x^3-15x+2=0
plz reply
Thank you sir this is very helpful
Sir plzz solution of binomial equation kese karte hai or innke properties ka video banao jaldi se....exam ke liye jada din nhi hai so sir plzz bsc 1st sem solutions of binomial equation me vedio banao
Amazing vídeo ..
Thank you :)
helped a lot
Whan t = -27 , -27 kaise aaya????
Good
Nice video bahut acha
THANK YOU SIR VERY INTERESTING I EASILY ABLE TO UNDERSTAND THE CONCEPT
First time I saw someone holding pen like me.....Happy ......😎😎😎
😃😂😂
Agr sir 2x^3 + x^2 + 1 ho qus tb isme missing term nhi hai pr x hai miss to isme me ese hoga solve jese apne kiya hai ki x^2 hi missing hona chahie ???
Put x = 1/y
Sir please solve x^3-12x+8=0
Hlo sir
Sir jo roots aege vo agl aa skte hai kya.... ???
Mere pas kuch or roots are hai or mene ye method lgaya hai book me kuch or trike se kiya hai usme kuch or roots ae hai
x3 - 12x - 65 =0 hai qus book me -1+3w or -1×3w2 hai ans ...or mere pas -5+ 3 undrrout 3 %2 ara hai ..plzz btaoo sir ...
Ur answer is correct.
I hv checked and 3rd root is 5.
Well done.
Aap aur bhi bsc ke liye video banaeye
Sir u and v have not one value, u and v have three values
Sir which device u have used for making this video and from where u got this
How in the world u took cube root of -27 as 3. Any root of negative number is imaginary number shouldn't it be -3j
Hlo sir. I am Arun. Can you give the answer of my question. My question is that X3 +X2 + X - 1 = 0. The mean of X3 = X cube, X2= X square plus X minus 1 = 0 .Are u able to give the solution of my question at date 28/04/18 to 29/04/18.
Why did you put the date?
Theory of equation wale pr bhi.video bnaye
If the value of (λ) is complex, Rational Numbers, Irrational Numbers
How can it I solve ?
Please make video
Example
(Q.1):x^4-4x^3+2x^2+12x=0
(Q.2):4x^4+2x^3-5x^2+6x-3=0
Please find roots by Ferrari's Method
awesome method to solve it
Sir, please Laplace transform and iska inverse ka bhi video bana dijiye and thanks .
Inverse of laplace transform-
ua-cam.com/video/bZnvrqjwMts/v-deo.html
Sir, thoda aur questions ka video daal dijiye please and thanks for it.
Root of negative- negative?
It's cube root
Negative ka qube kese nikalega Sir ji.. Ye to aayota ki from me aa jayega na
Abe sqrt me iota hota h
Sir I have a question. That is
3a^3+4a-3
Please solve this
Very good
Solve x^3-5x^2-16x+80=0 by carson's method please solve this question ❓🙏
Very good method
Thanks for liking
can you please solve the eq -x^3+3x+2=0 as If we put it in calci we get diff roots as compared to the one's that we get using this method
thanks
X³-3x-2=0
X³+x²-x²-x-2x-2=0
X²(x+1)-x(x+1)-2(x+1)=0
(X+1)(x²-x-2)=0
ans 1,1,2
@@bhaviias...370 putting 1 will not satisfy the given equation. Hence 1 can't be a possible root -1, -1 and 2 are the possible roots.
@@shashikamal5489 do This for me please.
x to 3rd power + x + 1 = O.
Watching this a day before exam
Sir Plss solve last question of cardan’s method
X2 ko kis trh htayengey ye bhi bta do
Thank You Very Much😘
You're welcome 😊
Thank's for this vidéo, please solve this équation : x³-7x-6=0 by cardan's méthode
This is the first video for solving cubic equation, which is without hit and trial 👏👏👍👍
Glad you liked it.
bahut bdiya
Very nice sir
Thanks a lot sir ❤
Superb
x^3 + x - 16 =0 give me solution plz sir ..........
Good explanation
How can u say 3 and 3 as two different roots ?
They came after solution of equation.
I must say u r 2 good
Good explaination.
Nice teaching method sir carry on brilliant class
Learning is a lifelong process mhh
Thanks sir