A) 1
B) 2
C) 3
D) 4
Correct Answer: B
Solution :
Here two cases arise viz. Case I : \[{{x}^{2}}+4x+3>0\] This gives \[{{x}^{2}}+4x+3+2x+5=0\] Þ\[{{x}^{2}}+6x+8=0\]Þ \[(x+2)(x+4)=0\] Þ \[x=-2,-4\] \[x=-2\] is not satisfying the condition \[{{x}^{2}}+4x+3>0\], so \[x=-4\] is the only solution of the given equation. Case II : \[{{x}^{2}}+4x+3<0\] This gives -\[({{x}^{2}}+4x+3)+2x+5=0\] Þ \[-{{x}^{2}}-2x+2=0\Rightarrow {{x}^{2}}+2x-2=0\] Þ\[(x+1+\sqrt{3})(x+1-\sqrt{3})=0\] Þ \[x=-1+\sqrt{3},-1-\sqrt{3}\] Hence \[x=-(1+\sqrt{3})\] satisfy the given condition\[{{x}^{2}}+4x+3<0\], while \[x=-1+\sqrt{3}\]is not satisfying the condition. Thus number of real solutions are two.
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