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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Answer:

Correct Answer: 50. ($x=-4$ and $(-1-\sqrt{3})$)

Solution:

  1. 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.



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