CARDAN'S METHOD- MATHEMATICS-2

This video will help to solve type 1 questions in cardon's method.

[niraj]

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

[arantheo]

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

[piya]

Smooth explaination, and very helpful for me, . Thank you🌹

[ar...parmar]

Wow sir ji bahut hi saral bhasa mai aapne kitni acchi taraha se is method ko samjha Diya bahut bahut dhanyavad ❤🔥🔥

[maintainedlife]

sir bahut hi ache tarike se smjaya apne pehli bar me hi dimag mai baith gya

[soumil]

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?

[davidbrisbane]

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).

[ahmedbaig]

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?

[thesurajgupta_]

Bhai mere khyal se (a+b)^3 ka formula a^3 + b^3 + 3(a^2)b + 3ab^2 hota hai.

[strikerstone]

This problem was too basic, easier with Rational root theorem, But i like this method 👍

[PriyaSingh-ulhl]

Your writing is amazing. Alsi the method

[jawedhussain]

What easy process to do this thnk u sir

[Abid_Ibn_Ashraf]

Sir, what if we get a irrational root and cannot perform synthetic division?

[ishantkatal]

Thanks sir I see ur previous videos of Ferrari n discarte understand ....all rules 😊😊😊😊😊

[attiqrehman]

Sir the Co efficient of x^3 should be 1 or if not then we have to take common the constant for comparison

[jawedhussain]

I want Ferrari method how to do on this simple trick pls make a video on that topic

[tarunmnair]

nice video, what if the complex roots of u and v are considered, then there are 9 solutions to x ?

[angelheretic]

How come did a cube become a quadratic? Can't understand, can someone explain.

[badaralam]

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

[patricemukulu]

Very helpful lesson but it was in English supposed to be more easier to understand but very helpful