Question

Solve $|x^2+4x+3|+2x+5=0$

Answer 1

Given, $\left|x^{2}+4 x+3\right|+2 x+5=0$
Case $\mid x^{2}+4 x+3>0 \Rightarrow(x<-3$ or $x>-1)$
$\therefore x^{2}+4 x+3+2 x+5=0$
$\Rightarrow x^{2}+6 x+8=0$
$\Rightarrow(x+4)(x+2)=0$
$\Rightarrow x=-4,-2[\text { but } x<-3 \text { or } x >-1]$
$\therefore x=-4$ is the only solution. ...(i)
Case II $x^{2}+4 x+3< 0 \Rightarrow(-3 < x < -1)$
$\therefore -x^{2}-4 x-3+2 x+5=0$
$\Rightarrow x^{2}+2 x-2=0 \Rightarrow(x+1)^{2}=3$
$\Rightarrow|x+1|=\sqrt{3}$
$\Rightarrow x=-1-\sqrt{3},-1+\sqrt{3}[\text { but } x \in(-3,-1)]$
$\therefore x=-1-\sqrt{3}$ is the only solution . ..(ii)
From Eqs. (i) and (ii), we get
$x=-4$ and $(-1-\sqrt{3})$ are the only solutions.