Question

If the roots of the equation `a(b − c)x^2 + b(c − a)x + c(a − b) = 0` are equal, then a, b, c are in ? (A) A.P. (B).G.P. (C) H.P. (D) none of these

Answer 1

(C) `a,b` and `c` are in H.P.

Explanation:

As the roots of `a(b-c)x^2+b(c-a)x+c(a-b)=0` are equal,

the discriminant of the equation is `0` i.e.

`b^2(c-a)^2=4ac(b-c)(a-b)`

or `b^2(c^2-2ac+a^2)=4ac(ab-ac+bc-b^2)`

or `b^2c^2-2acb^2+a^2b^2=4a^2bc-4a^2c^2+4abc^2-4acb^2`

or `4a^2c^2+a^2b^2-4a^2bc-4abc^2+2acb^2+b^2c^2=0`

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

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

Let `2c-b=t` then the above becomes

`a^2t^2-2abct+b^2c^2=0`

or `(at-bc)^2=0` i.e. `at-bc=0`

or `a(2c-b)-bc=0`

i.e. `2ac-ab-bc=0`

or `bc+ab=2ac`

or `bc-ac=ac-ab` and dividing by `abc`, we get

`1/a-1/b=1/b-1/c`

Hence, `a,b` and `c` are in H.P.

Answer 2

See below.

Explanation:

Solving for `x`

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

we have as roots

`x = {(1),(((a - b) c)/(a (b - c))):}`

or

`(a-b)c = a(b-c)` or dividing both sides by `abc`

`1/b-1/a=1/c-1/b rArr`HP