First slide

Introduction to Determinants

Question

If a > 0 and discriminant of ax² + 2bx + c is negative, then $Δ = |\begin{array}{ccc} a & b & ax + b \\ b & c & bx + c \\ ax + b & bx + c & 0 \end{array}|$ is

Difficult

A

+ve
B

$(ac - b)^{2}) ({ax}^{2} + 2 bx + c)$
C

-ve
D

0

Solution

Here a > 0 and 4b² - 4ac < 0, i.e., ac - b² > 0.
$∴ {ax}^{2} + 2 bx + c > 0, \forall x \in R$
Now,
$Δ = |\begin{array}{llc} a & b & ax + b \\ b & c & bx + c \\ 0 & 0 & - ({ax}^{2} + 2 bx + c) \end{array}| [[Operating R_{3} \to R_{3} - {xR}_{1} - R_{2}] = - ({ax}^{2} + 2 bx + c) (ac - b^{2}) = - (+ ve) (+ ve) = - ve$

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