First slide

Theory of expressions

Question

If the equation ${ax}^{2} + bx + c = 0, a, b, c \in R$ have non-real roots, then

Moderate

A

c(a - b + c) > 0
B

c(a + b + c) > 0
C

c(4a - 2b + c) > 0
D

none of these

Solution

Since the roots of ax² + bx + c = 0 are nonreal, so, f(x) = ax² + bx + c will have same sign for every value of x. Hence,
$\begin{array}{l} f (0) = c, f (1) = a + b + c, f (- 1) = a - b + c \\ f (- 2) = 4 a - 2 b + c \\ \Rightarrow c (a + b + c) > 0, c (a - b + c) > 0, c (4 a - 2 b + c) > 0 \end{array}$

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