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

Answer:

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,

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

Answer 2

Answer:

hope it helps you

Mark as BRAINLIEST