Question

If the roots of equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ be equal, then

• A. $2b=a+c$
• B. $b^2=ac$
• C. $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$
• D. None of these.

Answer 1

The correct option is C $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$

Given equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ ...... $\left(i\right)$

Let $\alpha ,\beta$ be the roots of the given quadratic equation $\left(i\right)$

$\Rightarrow \alpha +\beta =-\frac{b(c-a)}{a(b-c)}$ and $\alpha \beta =\frac{c(a-b)}{a(b-c)}$
Sum of coefficients $=0$
i.e $a(b-c)+b(c-a)+c(a-b)=0$

$\Rightarrow 1+\frac{b(c-a)}{a(b-c)}+\frac{c(a-b)}{a(b-c)}=0$

$\Rightarrow 1-\alpha -\beta +\alpha \beta =0$

$\Rightarrow 1-\alpha -\beta (1-\alpha )=0$

$\Rightarrow (1-\alpha )(1-\beta )=0$

$\Rightarrow 1-\alpha =0,1-\beta =0$

$\Rightarrow \alpha =1,\beta =1$

Then, product of the roots is

$\frac{c(a-b)}{a(b-c)}=1$

$\Rightarrow b=\frac{2ac}{a+c}$
Thus, $a,b,c$ are in H.P.
Hence, $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$.