Let a , b , c be in arithmetic progression. Let the centroid of the triangle with vertices a , c , 2 , b and…

Let a,b,c be in arithmetic progression. Let the centroid of the triangle with vertices a,c,2,b and a,b be 103,73. If α,β are the roots of the equation ax2+bx+1=0, then the value of α2+β2-αβ is:
  1. -71256
  2. 69256
  3. 71256
  4. -69256

Solution

a+2+a3=103

a=4

and c+b+b3=73

c+2b=7

also 2b=a+c

2b-a+2b=7

b=114

now 4x2+114x+1=0

 

α2+β2-αβ=α+β2-3αβ

=-11162-314

=121256-34=-71256

Asked in: JEE Main 2021 (24 Feb Shift 2)

Practice more Sequences and Series questions on Aicharya