A) Positive
B) \[(ac-{{b}^{2}})(a{{x}^{2}}+2bx+c)\]
C) Negative
D) 0
Correct Answer: C
Solution :
Let \[\Delta =\,\left| \,\begin{matrix} a & b & ax+b \\ b & c & bx+c \\ ax+b & bx+c & 0 \\ \end{matrix}\, \right|\] Applying \[{{R}_{3}}\to {{R}_{3}}-x{{R}_{1}}-{{R}_{2}};\] we get \[\Delta =\,\left| \,\begin{matrix} a & b & ax+b \\ b & c & bx+c \\ 0 & 0 & -(a{{x}^{2}}+2bx+c) \\ \end{matrix} \right|\,\] \[\Delta =({{b}^{2}}-ac)\,(a{{x}^{2}}+2bx+c)\] Now, \[{{b}^{2}}-ac<0\] and \[a>0\] Þ Discriminant of \[a{{x}^{2}}+2bx+c\] is -ve and \[a>0\] Þ \[(a{{x}^{2}}+2bx+c)>0\] for all \[x\in R\] Þ \[\Delta =({{b}^{2}}-ac)\,(a{{x}^{2}}+2bx+c)<0\], i.e.-ve.
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