Question

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

Answer 1

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

$\frac{\underset{\downarrow }{\underset{}{\left|(x+3)(x+1)\right|}}+2x+5=0+-+}{-1-3}$

$\to$ for

$x\in (-1,-3)$

$-(x^2+4x+3)+2x+5=0$

$x^2+4x+3-2x-5=0$

$x^2+2x-2=0$

$x=\frac{-2\pm \sqrt{4+8}}{2}$

$\to$ for

$x\in (-\infty ,-3)\cup [-1,\infty ]$

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

$x^2+6x+8=0$

$(x+4)(x+2)=0$

$x=-4,x=-2$

do not Lie

$(=\infty ,-3)\cup [-1,-\infty ]$

So,

$x=(-4)$ or $(-1-\sqrt{3})$