Question

If a,b,c are in HP, then the equation a(b-c)x²+ b(c-a)x +c(a-b) =0 has​

Answer 1

Answer:

say some easy question please

Step-by-step explanation:

are from which class

Answer 2

Step-by-step explanation:

If the roots are equal then the discriminant B^2 - 4AC=0

so b^2(c-a)^2 -4ac(b-c)(a-b)=0

This gives

a^2b^2+b^2c^2+4a^2c^2 +2ab^2c-4a^2bc-4abc^2=0

This can be expressed as (ab+bc)^2 -4abc(a+c)+4a^2c^2=0

or b^2(a+c)^2 - 4abc(a+c)+4a^2c^2=0

This is a perfect square,

[b(a+c) -2ac]^2=0 and so

b(a+c) - 2ac=0 and assuming that none of a,b or c =0,

dividing by abc gives 1/c+1/a - 2/b=0 and hence

2/b=1/a+1/c