First slide

Theory of equations

Question

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

Moderate

A

$b < 0 < a$
B

$a < 0 < b$
C

$b < a < 0$
D

None of these

Solution

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

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