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

$b > a > 0$

Solution

We have, $D = (b - c)^{2} - 4 a (a - b - c) > 0$
$\begin{array}{l} \Rightarrow b^{2} + c^{2} - 2 b c - 4 a^{2} + 4 a b + 4 a c > 0 \\ \Rightarrow c^{2} + (4 a - 2 b) c - 4 a^{2} + 4 a b + b^{2} > 0, \forall c \in R \end{array}$
Since, $c \in R$ , so we have
$\begin{array}{l} (4 a - 2 b)^{2} - 4 (- 4 a^{2} + 4 a b + b^{2}) < 0 \\ \Rightarrow 4 a^{2} - 4 a b + b^{2} + 4 a^{2} - 4 a b - b^{2} < 0 \\ \Rightarrow a (a - b) < 0 \end{array}$
If $a > 0,$ then $a - b < 0$
i.e. $0 < a < b$
or $b > a > 0$
If $a < 0, then a - b > 0$
i.e. $0 > a > b$
or $b < a < 0$

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