In Descarte's method we can multiply quadratics and compare coefficients even without getting rid of x^3 term We will simply substitute p=y+h in the resolvent sextic
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You are explain very well but in 7:07 some mistake.... q-q'=-24/p
Not...q-q' =24/q.....
But explanation is very well mam
Thanks💝💝
In Descarte's method we can multiply quadratics and compare coefficients even without getting rid of x^3 term
We will simply substitute p=y+h in the resolvent sextic
Thankyou ma'm, you explained very well
Please make a video with third term coefficient in quadratic equation with descartes method
If x^3 term is present you should get rid of it by substitution
otherwise computations complicate a lot
Thanks 🙏🙏
Well explained 👍
Thanks mam
Thanks ma'am
Plz make a video at solve the biquadratic equation by Euler's method.🙏🙏
Sure
It is generalization of Cardan's method for cubics
Arup Majumdar recorded video for it
Good
Gajab he
Thankuu so much mam🙏🙏
Solve x^4-10x^2-20x-10=0 by descartes method ...... solve this prblm
Your squaring part is complicated
🙏😊
Please mam eular method
ua-cam.com/video/SiDzBpslfT8/v-deo.html
ua-cam.com/video/fDFUFCXjFB8/v-deo.html
Tnx mam
Thnks
Tq mam🙏
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Ganesh Institute!!
No no don't make any new videos mam it's too horrible
I can't understand u r class
Thanks mam
Thanks mam