Theory of Equations 1 Question 49
50. Solve $\left|x^{2}+4 x+3\right|+2 x+5=0$
(1987, 5M)
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Solution:
- Given, $\left|x^{2}+4 x+3\right|+2 x+5=0$
Case I $\quad x^{2}+4 x+3>0 \Rightarrow(x<-3$ or $x>-1)$
$\therefore \quad x^{2}+4 x+3+2 x+5=0$
$$ \Rightarrow \quad x^{2}+6 x+8=0 \Rightarrow(x+4)(x+2)=0 $$
$\Rightarrow \quad x=-4,-2 \quad$ [but $x<-3$ or $x>-1$ ]
$\therefore \quad x=-4$ is the only solution.
Case II $\quad x^{2}+4 x+3<0 \Rightarrow(-3<x<-1)$
$\therefore \quad-x^{2}-4 x-3+2 x+5=0$
$$ \Rightarrow \quad x^{2}+2 x-2=0 \Rightarrow(x+1)^{2}=3 $$
$\Rightarrow \quad|x+1|=\sqrt{3}$
$\Rightarrow \quad x=-1-\sqrt{3},-1+\sqrt{3} \quad$ [but $x \in(-3,-1)]$
$\therefore \quad x=-1-\sqrt{3}$ is the only solution.
From Eqs. (i) and (ii), we get
$x=-4$ and $(-1-\sqrt{3)}$ are the only solutions.
