Question

If a(b-c) x^2 + b(c-a) x + c(a-b) = 0 has equal roots, show that 2/b = 1/a + 1/c​

Answer 1

$\large\underline{\sf{Solution-}}$

Given quadratic equation is

$\rm :\longmapsto\:a(b - c) {x}^{2} + b(c - a)x + c(a - b) = 0$

Concept Used :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac

So, On comparing with Ax² + Bx + C = 0, we get

$\red{\rm :\longmapsto\:A = a(b - c)}$

$\red{\rm :\longmapsto\:B = b(c - a)}$

$\red{\rm :\longmapsto\:C = c(a - b)}$

So, According to statement, equation have real and equal roots.

So, Discriminant = 0

$\rm \implies\: {B}^{2} - 4AC = 0$

$\sf \: {\bigg[b(c - a)\bigg]}^{2} - 4a(b - c)c(a - b) = 0$

$\sf \: {b}^{2} {(c - a)}^{2} - 4ac(ab - ca - {b}^{2} + bc) = 0$

$\sf \: {b}^{2}[ {c}^{2}+{a}^{2} - 2ac] - 4ac(ab - ca - {b}^{2} + bc) = 0$

$\sf \: {b}^{2}{c}^{2}+ {b}^{2}{a}^{2} - 2ac {b}^{2}- 4cb {a}^{2} + 4 {a}^{2}{c}^{2} + 4ac{b}^{2} - 4ab {c}^{2}= 0$

$\sf \:{b}^{2}{c}^{2}+ {b}^{2}{a}^{2} + 2ac {b}^{2}- 4cb {a}^{2} + 4 {a}^{2}{c}^{2} - 4ab {c}^{2}= 0$

$\sf \:{b}^{2}{c}^{2}+ {b}^{2}{a}^{2} + {4a}^{2} {c}^{2} + 2ac {b}^{2}- 4cb {a}^{2} - 4ab {c}^{2}= 0$

$\rm \implies\: {(ab + bc - 2ac)}^{2} = 0$

$\rm \implies\: ab + bc - 2ac = 0$

$\rm \implies\: ab + bc = 2ac$

On dividing both sides by abc, we get

$\rm \implies\:\dfrac{ab}{abc} + \dfrac{bc}{abc} = \dfrac{2ac}{abc}$

$\bf \implies\:\dfrac{1}{c} + \dfrac{1}{a} = \dfrac{2}{b}$

Hence, Proved