That's What I'm Talking About!

We have, $\left| { x }^{ 2 }+4x+3 \right| +2x+5$

Here two cases arise.

• Case 1:- When,

$\Rightarrow$ ${ x }^{ 2 }+4x+3>0$

$\Rightarrow$ ${ x }^{ 2 }+4x+3+2x+5=0$

$\Rightarrow$ ${ x }^{ 2 }+6x+8=0$

$\Rightarrow$ Hence, x=-2, -4

Here, x=-2 doesn't satisfy ${ x }^{ 2 }+4x+3>0$ . So, x=-4 is the only solution for Case 1!

• Case 2:- When,

$\Rightarrow$ ${ x }^{ 2 }+4x+3<0$

$\Rightarrow$ $-{ x }^{ 2 }-4x-3+2x+5=0$

$\Rightarrow$ $-{ x }^{ 2 }-2x+2=0$

$\Rightarrow$ ${ x }^{ 2 }+2x-2=0$

$\Rightarrow$ $x=-1+\sqrt { 3 } ,-1-\sqrt { 3 }$

$\Rightarrow$ Hence, $x=-(1+\sqrt { 3 })$ satisfies the given condition, while, $x=-1+\sqrt { 3 }$ doesn't!

• From Case 1 & Case 2 , we get 2 real roots!

We have two cases: 1. When $-x^2-4x-3+2x+5=0$

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

$x=-1 \pm \sqrt{3}$

1. When $x^2+4x+3+2x+5=0$

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

$x=-2,-4$

Sustitute in the original equation only $-2$ and $-1-\sqrt{3}$ are solutions.