Correct answer is : 2
\(|x+1||x+3|-4|x+2|+5=0\)
case- 1
\(\mathrm{x} \leq-3\)
\((\mathrm{x}+1)(\mathrm{x}+3)+4(\mathrm{x}+2)+5=0\)
\(\mathrm{x}^{2}+4 \mathrm{x}+3+4 \mathrm{x}+8+5=0\)
\(\mathrm{x}^{2}+8 \mathrm{x}+16=0\)
\((\mathrm{x}+4)^{2}=0\)
\(\mathrm{x}=-4\)
case-2
\(-3 \leq \mathrm{x} \leq-2\)
\(-\mathrm{x}^{2}-4 \mathrm{x}-3+4 \mathrm{x}+8+5=0\)
\(-\mathrm{x}^{2}+10=0\)
\(\mathrm{x}= \pm \sqrt{10}\)
case-3
\(-2 \leq \mathrm{x} \leq-1\)
\(-x^{2}-4 x-3-4 x-8+5=0\)
\(-x^{2}-8 x-6=0\)
\(\mathrm{x}^{2}+8 \mathrm{x}+6=0\)
\(x=\frac{-8 \pm 2 \sqrt{10}}{2}=-4 \pm \sqrt{10}\)
case-4
\(\mathrm{x} \geq-1\)
\(x^{2}+4 x+3-4 x-8+5=0\)
\(\mathrm{x}^{2}=0\)
\(\mathrm{x}=0\)
No. of solution = 2
