Question

If $a,b,c$ are in H.P., then the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$

• A. has real and distinct roots.
• B. has equal roots.
• C. has no real roots.
• D. has 1 as a root.

Answer 1

Answer: (B), (D)

The correct options are
B has equal roots.
D has 1 as a root.

Substituting $x=1$, we get
$ab-ac+bc-ab+ac-bc$$=0$

Hence, $x=1$ is a root.

$\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P. As a,b,c are in H.P.

Hence,$\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$Or $\frac{1}{b}-\frac{1}{c}=\frac{1}{a}-\frac{1}{b}$.

$\frac{c-b}{bc}=\frac{b-a}{ab}$.

$a(c-b)=c(b-a)\dots \dots \left(i\right)$

The standard quadratic equation is $m{x}^2+nx+k=0$

and Product of roots = $\frac{k}{m}$

Here, Product of roots $=\frac{c(a-b)}{a(b-c)}$.

$=\frac{c(b-a)}{a(c-b)}$.

$=1\dots \dots$ from $\left(i\right)$.

Hence, product of roots is $1$.
Therefore, both roots are $1$.