First slide

Theory of equations

Question

If ${ax}^{2} + (b - c) x + a - b - c = 0$ has unequal real roots for all $c \in R$ , then

Moderate

A

b < 0 < a
B

a < 0 < b
C

b < a < 0
D

b > a > 0

Solution

We have,
$\begin{array}{l} D = (b - c)^{2} - 4 a (a - b - c) > 0 \\ or b^{2} + c^{2} - 2 bc - 4 a^{2} + 4 ab + 4 ac > 0 \\ or c^{2} + (4 a - 2 b) c - 4 a^{2} + 4 ab + b^{2} > 0 for all c \in R \end{array}$
Discriminant of the above expression in c must be negative.
Hence,
$\begin{array}{ll} (4 a - 2 b)^{2} - 4 (- 4 a^{2} + 4 ab + b^{2}) < 0 \\ or 4 a^{2} - 4 ab + b^{2} + 4 a^{2} - 4 ab - b^{2} < 0 \\ or a (a - b) < 0 \\ \Rightarrow a < 0 and a - b > 0 or a > 0 and a - b < 0 \\ \Rightarrow b < a < 0 or b > a > 0 \end{array}$

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