First slide

Theory of equations

Question

If both roots of the equation ${ax}^{2} + x + c - a = 0$ are imaginary and c > -1, Then

Easy

A

$3 a > 2 + 4 c$
B

$3 a < 2 + 4 c$
C

c < a
D

none of these

Solution

Let,
$\begin{array}{l} f (x) = {ax}^{2} + x + c - a \\ f (1) = c + 1 > 0 (∵ c > - 1) \end{array}$
Therefore given expression is positive $\forall x \in R$ . So,
$f (\frac{1}{2}) > 0$
$\Rightarrow \frac{a}{4} + \frac{1}{2} + c - a > 0$
$\begin{array}{l} or 4 c - 3 a + 2 > 0 \\ or 4 c + 2 > 3 a \end{array}$

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