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If $a > 0$and discriminant of $a{x^2} + 2bx + c$is negative, then $\left| {\,\begin{array}{*{20}{c}}a&b&{ax + b}\\b&c&{bx + c}\\{ax + b}&{bx + c}&0\end{array}\,} \right|$ is
Positive
$(ac - {b^2})(a{x^2} + 2bx + c)$
Negative
$0$
Solution
(c) Let $\Delta = \,\left| {\,\begin{array}{*{20}{c}}a&b&{ax + b}\\b&c&{bx + c}\\{ax + b}&{bx + c}&0\end{array}\,} \right|$
Applying ${R_3} \to {R_3} – x{R_1} – {R_2};$ we get
$\Delta = \,\left| {\,\begin{array}{*{20}{c}}a&b&{ax + b}\\b&c&{bx + c}\\0&0&{ – (a{x^2} + 2bx + c)}\end{array}} \right|\,$
$\Delta = ({b^2} – ac)\,(a{x^2} + 2bx + c)$
Now, ${b^2} – ac < 0$ and $a > 0$
==> Discriminant of $a{x^2} + 2bx + c$ is -ve and $a > 0$
==> $(a{x^2} + 2bx + c) > 0$ for all $x \in R$
==> $\Delta = ({b^2} – ac)\,(a{x^2} + 2bx + c) < 0$, i.e.-ve.
