@4:57 You have z(2z^2+3/2)=0 z=0 or 2z^2+3/2=0 Solve for z: z=0 or z=± {sqrt(3)i }/4 If z=x+1/2 , and z=0 or ± {sqrt(3)i }/4; Solve for x': 0=x+1/2 x'=-1/2 (1st root) Solve for x'' and x''': ± {sqrt(3)i }/4=x+1/2 x''=-1/2 + {sqrt(3i) }/4 (2nd root) x'''=-1/2 - {sqrt(3i) }/4 (3rd root)
Sir can't we don't in start like formula of my mam H= - of a1/ highest power I think it will be easier with same ans Plz reply Bez I have don't sums with same method
@@civiluptodate1580 sir the method you applied is very efficient but when the value of roots comes as 5/4 or any in fraction form it is difficult to understand
because the method is only eligible of solving a third degree equation thus you can use the regular formula of 2nd degree equation afterwards to find the two remaining answers
Carden's method is not something that you change your equation by setting the coefficient of (n-1)th term to zero. What carden' s method does is that it shifts the whole function to make it standard, same as we do in a quadratic equation and then solve it. In quadratic equation, we also change our x by (x-b/2a) to make it in a perfact square form and then solve it same as cardern does.
2x³-15x²+24x-7 sir nahi nikalta jeh sum plz help sir kindly reply as soon as possible Sir h ata hai mera 5/2 pr sir joh Mr. Seven hai voh cancel nahi hota I tried 7/2 also but result remains same So plz sir reply and help me
@@civiluptodate1580 It is easier to solve it by writing this polynomial as x^3+(x+1)^3 but maybe there is something in giving examples which can be solved easy with other method
U r just awesome ☺️ sir....now I have some confidence that I will attempt these questions and will pass in
Good explanation dear bro
Tnx sir you are such a good teacher
You can that équation equal to x^3+ (x+1)^3 and to continue .
yeah i did the same , way easier method
Thank you so much sir. It helped me . 👍
@4:57 You have z(2z^2+3/2)=0
z=0 or 2z^2+3/2=0
Solve for z:
z=0 or
z=± {sqrt(3)i }/4
If z=x+1/2
, and z=0 or ± {sqrt(3)i }/4;
Solve for x':
0=x+1/2
x'=-1/2 (1st root)
Solve for x'' and x''':
± {sqrt(3)i }/4=x+1/2
x''=-1/2 + {sqrt(3i) }/4
(2nd root)
x'''=-1/2 - {sqrt(3i) }/4 (3rd root)
Ap n leading coefficient remove ni kiya.q?
1st step m to hm leading coefficient remove krty h na
Thanking you
Good but I don't understand why I divide x+1/2
Jab h=1/2 se divide kiya tha to reminder 0 aya to 1 soln wahi mil gya tha ...to agey synthetic ko increase ku kiya..
sudhir trivedi x² term eliminate krne k liye
i did this in a few mins w easier method , i knew what to do when a cubic has 3x^2+3x in the equation
Sir isme apne -1/2 se divide kiya thaa question ko toh ek eq.s ayi jisme constant zero hua
Fir uske baad hamne solve kiya toh x= -1/2 , ek root ki value agayi ,
Pr uske baad firse kyu divide kiya ques wali eq. ko aur tab alag eq kyunki aapne jab divide krte time jaisi 2. 2. 2. 0. Toh apne yahi rok diya aur divide nahi kiyu uske baad
Aisa kyu?
Great sir
U really teach good. Thanks sir
2nd method जिसमे 2 times karege?????????????????????
very nice sir 👏👏👏👏
Sir can't we don't in start like formula of my mam
H= - of a1/ highest power
I think it will be easier with same ans
Plz reply
Bez I have don't sums with same method
I think in your case value of a0 is one.
@@civiluptodate1580 in some question sir but not all
@@civiluptodate1580 sir the method you applied is very efficient but when the value of roots comes as 5/4 or any in fraction form it is difficult to understand
@@civiluptodate1580 sir like 5his question use Ferrari method for
X^4-8x^3+11x^2+20x+4 =0
@@civiluptodate1580 here I am unable to know how to find root in rhs while making it complete sq as it's root comes -5/2
Nice
What is the formula for h removal of third term
Sir I have a question...why cardan's method couldn't solve 2nd degree eqation
because the method is only eligible of solving a third degree equation thus you can use the regular formula of 2nd degree equation afterwards to find the two remaining answers
Carden's method is not something that you change your equation by setting the coefficient of (n-1)th term to zero. What carden' s method does is that it shifts the whole function to make it standard, same as we do in a quadratic equation and then solve it. In quadratic equation, we also change our x by (x-b/2a) to make it in a perfact square form and then solve it same as cardern does.
Thankyou
Sir iss question ka solution aap bta skte h kya X3-3X2+3=0 x k sth 3 aur 2 power m h cardan's method ka h
Put x = 1/y
@@civiluptodate1580 sir how to find roots of this cubic equation 4x^3-3x^2+2x-1=0 response plzzzz
It was helpful
Sir b.sc part 1 ka math full solution
2x^2-3x+1=0 cardan method please solve it..
Thank you Sir
2x³-15x²+24x-7 sir nahi nikalta jeh sum plz help sir kindly reply as soon as possible
Sir h ata hai mera 5/2 pr sir joh Mr. Seven hai voh cancel nahi hota
I tried 7/2 also but result remains same
So plz sir reply and help me
Thanku sir
Thanks sir
thanx
5:11 isme apn z ki ek or value b nikaal k put ni kr Skte kya
3:57 se smjh ni aaya 5:10 tkk , help 💛
Sir how you record your vedios
sir one question discarte method ka
Camera ko stand kaise kia h
By placing on the edge of table & writing on lower table
Civil Up to date Thankyou so much sir
Can we use this method in solving higher degree polynomials sir?
No
Sir image nary root ka power 1/3 banega toh solve methods bathae
Sir agr -1/2 ki jga 6/7 ho to sol k hoga
aap i ko iota kyo nhi bolte ??
Jis method say saray karwaye thay usi method say yeh bhi karwa deta bhai 😤 sab ka method alag alag seekhay gay kya
Very fast
Example is not well chosen
See my other two videos
@@civiluptodate1580 It is easier to solve it by writing this polynomial as x^3+(x+1)^3
but maybe there is something in giving examples which can be solved easy with other method
Sir can't you speak english?
Sie g samj ma nhi aaa rha
See my other two videos
U make a mistake at 2:43
There is no mistake.
U please check again.
@@civiluptodate1580 2+1/2=3/2
There is (-1/2).
Check again.
samag me nahi aaya
Which step ?