Question

If the roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are equal then prove that $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$ i.e. $a,b,c$ are in H.P.

Answer 1

Given quadratic equation,

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

Zeros of the given equation are equal.

So, discriminant $=0$

$b^2-4ac=0$

$\left[b\right(c-a){]}^2-4[a(b-c)\cdot c(a-b)]=0$

$b^2(c^2+a^2-2ac)-4\left[\right(ab-ac\left)\right(ac-cd\left)\right]=0$

$b^2{c}^2+a^2{b}^2-2a{b}^{2}c-4[a^{2}bc-a{b}^{2}c-a^2{c}^2+ab{c}^2]=0$

$b^2{c}^2+a^2{b}^2+2a{b}^c-4a^{2}bc+4a{b}^{2}c+4a^2{c}^2-4ab{c}^2=0$

$b^2{c}^2+a^2{b}^2+2a{b}^{2}c-4a^{2}bc+4a^2{c}^2-4ab{c}^2=0$

We know that,

$a^2+b^2+c^2+2ab+2bc+2ca=(a+b+c{)}^2$

By above identity we get,

$[bc+ab-2ac{]}^2=0$

$bc+ab-2ac=0$

$b(c+a)=2ac$

$\frac{c+a}{ac}=\frac{2}{b}$

$\Rightarrow \frac{c}{ac}+\frac{a}{ac}=\frac{2}{b}$

$\therefore \frac{2}{b}=\frac{1}{a}+\frac{1}{c}$.