A) 4
B) 3
C) 2
D) 1
Correct Answer: C
Solution :
Equation is \[|3{{x}^{2}}+12x+6|\,=5x+16\] ?..(i) when \[3{{x}^{2}}+12x+6\ge 0\]\[\Leftrightarrow \,\,\,{{x}^{2}}+4x\ge -2\] \[\Leftrightarrow \,\,\,|x+2{{|}^{2}}\ge 4-2\,\,\,\Leftrightarrow \,\,\,|x+2|\,\ge {{(\sqrt{2})}^{2}}\] \[\Leftrightarrow \,\,\,x+2\le -\sqrt{2}\]or \[x+2\ge \sqrt{2}\] ....(ii) Then (i) becomes \[3{{x}^{2}}+12x+6=5x+16\] \[\Leftrightarrow \,\,\,3{{x}^{2}}+7x-10=0\Rightarrow x=1,\,-\frac{10}{3}\] But \[x=-\frac{10}{3}\] does not satisfy (ii). When \[3{{x}^{2}}+12x+6<0\]\[\Rightarrow {{x}^{2}}+4x<-2\] Þ \[\,|x+2|\,\le \sqrt{2}\]\[\Rightarrow \,\,-\sqrt{2}-2\le x\le -2+\sqrt{2}\] ?..(iii) Then (i) becomes \[\Rightarrow 3{{x}^{2}}+12x+6=-(5x+16)\] \[\Rightarrow \,\,\,\,3{{x}^{2}}+17x+22=0\Rightarrow x=-2,-\frac{11}{3}\] But \[x=-\frac{11}{3}\]does not satisfy (iii). So, 1 and - 2 are the only solutions.
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