Question

The real solutions of $|x^2+4x+3|+2x+5=0$ is/are

• A. $-1-\sqrt{3}$
• B. $-1+\sqrt{3}$
• C. $-2$
• D. $-4$

Answer 1

Answer: (A), (D)

The correct options are
A $-1-\sqrt{3}$
D $-4$

The Given equation is, $|x^2+4x+3|+2x+5=0$
Now there can be two cases.
Case $1$:
$x^2+4x+3\ge 0\Rightarrow (x+1)(x+3)\ge 0$
$\Rightarrow x\in (-\infty ,-3]\cup [-1,\infty )\cdots \left(1\right)$
Then given equation becomes,
$x^2+4x+3+2x+5=0$
$\Rightarrow x^2+6x+8=0$
$\Rightarrow (x+4)(x+2)=0\Rightarrow x=-4,-2$
But $x=-2$ does not satisfy $\left(1\right)$, hence rejected
$\therefore x=-4$ is the solution.

Case $2$ :
$x^2+4x+3<0$
$\Rightarrow (x+1)(x+3)<0$
$\Rightarrow x\in (-3,-1)\cdots \left(2\right)$
Then given equation becomes,
$-(x^2+4x+3)+2x+5=0$
$\Rightarrow -x^2-2x+2=0\Rightarrow x^2+2x-2=0$
$\Rightarrow x=\frac{-2\pm \sqrt{4+8}}{2}\Rightarrow x=-1+\sqrt{3},-1-\sqrt{3}$
Out of which $x=-1-\sqrt{3}$ is solution.
Combining the two cases we get the solutions of given equation as $x=-4,-1-\sqrt{3}$