First slide

Theory of expressions

Question

If roots of the equation (a - b)x² + (c - a)x+(b - c) = 0 are equal, then a, b and c are in

Easy

A

AP
B

HP
C

GP
D

None of these

Solution

Since, roots of the equation
$(a - b) x^{2} + (c - a) x + (b - c) = 0$
$∴ Discriminant, B^{2} - 4 AC = 0$
$\begin{array}{lr} \Rightarrow (c - a)^{2} - 4 (a - b) (b - c) = 0 \\ \Rightarrow a^{2} + 4 b^{2} + c^{2} + 2 ac - 4 ab - 4 bc = 0 \\ \Rightarrow (a + c - 2 b)^{2} = 0 \Rightarrow a + c = 2 b \end{array}$
Hence ,a, b, c ate in AP

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App