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: 11

                        

Let a,b be arbitary real numbers. Find the smallest natural number `b` for which the equation x^2+2 (a+b)x+(a-b+8)=0 has unequal real roots for all a belongs to real numbers.

2 years ago

Answers : (2)

Ritesh Khatri
76 Points
							(a-b+8) > 0If D 2 – 4
a2 + b2 + 2ab – a – b – 8 >  0
a2 + (2b – 1)a + b2 – 8 >  0
it is given a belongs to real numbers and abobe quadratic is always greater than 0 hence polynomial a2 + (2b – 1)a + b2 – 8 can not have real roots 
hence its D
(2b – 1)2 – 4(b2 – 8) 
 
- 4b + 1 + 24 
b > 25/4
 
smallest naturan number b = 7
 
2 years ago
Ritesh Khatri
76 Points
							
Initial line is for unequal roots discriminant of given equation must be greater than 0, D > 0 
(a-b+8) > 0(2 (a+b))^2 – 4
 
2 years 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