@@Sakshijonwal2006, almost like the video we did beta. Let the roots be a, a+b and a+ 2b. Sum of the roots is -b/a So, a+a+b+a+2b = 36/8 = 9/2 Therefore a+b = 9/6 = 3/2. We also know that product of roots is -d/a = -81/8. So, a(a+b)(a+2b) = -81/8 (3/2-b)(3/2/(3/2+b) = -81/8 9/4 - b^2 = -27/4 So, b^2 = 9 or b = plus or minus 3. So, the numbers are -3/2, 3/2 and 9/2. Please let me know if this solution helped you. 👍😊❤️
@@RajendraMahara-g5f 9th grade solution: P(1) = 0 beta. So, x-1 is a factor. Now divide p(x) by x-1 . You get x^2-11x+28. Factors of this quadratic expression are (x-4) and (x-7). So, the roots are 1,4 and 7. 11th grade solution. ax^3+ bx^2+ cx + d= 0 x^3-12x^2 +39x -28=0 Sum of roots = -b/a Alpha+ beta+ gamma = 12/1 = 12 Sum of product of roots taken two at a time = c/a Alpha*beta + beta*gamma+ gamma*Alpha = 39/1 = 39 Product of the roots = -d/a Alpha* beta * gamma = 28/1 = 28.
Thank u sir good explanation this video is very useful for me 😊
Can we put a,a+b,a+2b as zeroes of f polynomial?
Yes.
Can you sir make video on these value
@@Sakshijonwal2006, almost like the video we did beta.
Let the roots be a, a+b and a+ 2b.
Sum of the roots is -b/a
So, a+a+b+a+2b = 36/8 = 9/2
Therefore a+b = 9/6 = 3/2.
We also know that product of roots is -d/a = -81/8.
So, a(a+b)(a+2b) = -81/8
(3/2-b)(3/2/(3/2+b) = -81/8
9/4 - b^2 = -27/4
So, b^2 = 9 or b = plus or minus 3.
So, the numbers are
-3/2, 3/2 and 9/2.
Please let me know if this solution helped you.
👍😊❤️
Thankyou sir this helped me a lot😁
Sir plss tell the name of book from Which u took the equation
Do not remember exactly beta, may be from higher algebra by hall and knight.
good explaination sir..
Today is my exam and iam studying this now...
All the best ☺❤👍🙏
Sir theory of equation most important questions say sir plzz for board 20233
2023
Thank you so much sir
The answer is - -2, 1,4
Is this correct?
Thank you so much brother or sir
Sir please tell the roots of this equation..
x^3-12x^2+39x-28=0
@@RajendraMahara-g5f
9th grade solution:
P(1) = 0 beta. So, x-1 is a factor.
Now divide p(x) by x-1 .
You get x^2-11x+28. Factors of this quadratic expression are (x-4) and (x-7).
So, the roots are 1,4 and 7.
11th grade solution.
ax^3+ bx^2+ cx + d= 0
x^3-12x^2 +39x -28=0
Sum of roots = -b/a
Alpha+ beta+ gamma = 12/1 = 12
Sum of product of roots taken two at a time = c/a
Alpha*beta + beta*gamma+ gamma*Alpha = 39/1 = 39
Product of the roots = -d/a
Alpha* beta * gamma = 28/1 = 28.
Thanks sir
9/4-d^2=-27/4 how to solve
Reply
Which class you are in beta ?
This video is for 12th grade students.
Sir nuv thop kali
👍😊❤️🙏
Hi
Hi :)
Thank u sir good explanation this video is very useful for me 😊