MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 12
        
If a, b, c are the roots of x3-2x2+6x-1=0, then find the value of ∑(a3+a2+a)/(a2-a+1).
11 months ago

Answers : (2)

Arun
22545 Points
							
Dear student
 
if we find f’(x)
then we know that it is always increasing fn hence it has only one real root
 
 
11 months ago
saket
12 Points
							
we have, (a3+a2+a)/(a2-a+1) = a(a2+a+1)/(a2-a+1) = a(a3-1)/{(a2-a+1)(a-1)} = a(a3-1)/(a3-2a2+2a-1)
= a(a3-1)/(a3-2a2+6a-1-4a) = (1-a3)/4,               [ since ‘a’ is a root of x3-2x2+6x-1, a3-2a2+6a-1=0 ]
similarly (b3+b2+b)/(b2-b+1) = (1-b3)/4 and 
              (c3+c2+c)/(c2-c+1) = (1-c3)/4.
therefore the required sum is {3-(a3+b3+c3)}/4.
now,
       a3+b3+c3-3abc = (a+b+c){(a+b+c)2-3(ab+bc+ca)}
also a+b+c=2, ab+bc+ca=6, and abc=1.
hence a3+b3+c= -25.
therefore the sum is {3-(-25)}/4 = 7.
 
11 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details