Question

If the roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are equal, show that $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$

Answer 1

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

Since the roots are equal, the discriminant is equal to $0$.

$\Rightarrow b^2(c-a{)}^2-4ac(b-c\left)\right(a-b)=0$

$\Rightarrow b^2(c^2+a^2-2ac)-4ac(ba-b^2-ca+bc)=0$

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

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

$\Rightarrow b^2(a^2+c^2+2ac)-4abc(a+c)+4a^2{c}^2=0$

$\Rightarrow b^2(a+c{)}^2-4abc(a+c)+4a^2{c}^2$

The above is a quadratic equation involving $(a+c)$

The roots of the equation is given by

$a+c=\frac{4abc\pm \sqrt{16a^2{b}^2{c}^2-16a^2{b}^2{c}^2}}{2b^2}$

$=\frac{4abc}{2b^2}=\frac{2ac}{b}$

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

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

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

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