Question

If
the roots of the equation (a -b) x2 + (b-c) x +
(c-a)=o are equal, show that c, a
and b are in AP.​

Answer 1

Given expression,

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

The above equation has equal roots.

$\sf Discriminant = 0$

For any three numbers to be in AP, the condition is :

$\sf \: 2z = x + y$

Now,

$\sf Discriminant = 0 \\ \\ \implies \sf \: B^2 - 4AC = 0 \\ \\ \implies \sf \: (b - c) {}^{2} - 4(a - b)(c - a) = 0 \\ \\ \implies \sf \: b {}^{2} + {c}^{2} - 2bc = 4ac \: - 4bc - \: 4ab - 4a {}^{2} \\ \\ \implies \sf \: (2a - b - c) {}^{2} = 0 \\ \\ \implies \boxed{ \boxed{ \sf 2a = b + c }}$

Answer 2

Required Answer :-

We may observe that the given equation has equal roots.

So,

Here

Apply discriminant formula

$\bf b^{2} - 4ac = 0$

Here,

$\bf b = ( b -c)$

$\bf a = (a - b)$

$\bf c = (c-a)$

$\tt(b - c)^{2}-4(a-b)(c-a)$

Apply identity

$\tt (a - b)^{2} = a^{2} - 2ab + b^{2}$

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

$\tt b^{2} + c^{2} - 2bc = ac - bc - ab - a^{2}$

$\tt (2a - b - c)^{2} = 0$

$\tt 2a = - b + - c$

$\tt 2a = b +c$