Question

solve the problem with solution ​

Answer 1

Answer:

given roots are equal

a(b-c)x^2 + b(c-a)x + c(a-b)= 0

abx^2 - acx^2 + bcx - abx + ac - bc = 0

ab(x^2 - x) + bc( x - 1) - ac( x^2 - 1) = 0

(x - 1)[ abx + bc - ac (x + 1)] = 0

this gives x - 1 = 0

x = 1

or

abx + bc - ac(x+1) = 0

since roots are same

x = 1 is the only solution

putting x= 1 , we get

ab + bc - 2ac = 0

b(a+c) = 2ac

b = 2ac/(a+c)

hence, a, b, c are in HP

Options b and c are correct