The number of real solutions of the equation | vedclass

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(B) The given equation is $|x^2 + 4x + 3| + 2x + 5 = 0$.Case $I$: $x^2 + 4x + 3 \ge 0$,which implies $(x+1)(x+3) \ge 0$,so $x \in (-\infty, -3] \cup [

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$2$

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(B) The given equation is $|x^2 + 4x + 3| + 2x + 5 = 0$.Case $I$: $x^2 + 4x + 3 \ge 0$,which implies $(x+1)(x+3) \ge 0$,so $x \in (-\infty, -3] \cup [-1, \infty)$.The equation becomes $x^2 + 4x + 3 + 2x + 5 = 0$,which simplifies to $x^2 + 6x + 8 = 0$.Factoring gives $(x+2)(x+4) = 0$,so $x = -2$ or $x = -4$.Checking the condition: $x = -4$ satisfies $x \in (-\infty, -3]$,but $x = -2$ does not satisfy $x \in [-1, \infty)$. Thus,$x = -4$ is a solution.Case $II$: $x^2 + 4x + 3 The equation becomes $-(x^2 + 4x + 3) + 2x + 5 = 0$,which simplifies to $-x^2 - 2x + 2 = 0$,or $x^2 + 2x - 2 = 0$.Using the quadratic formula,$x = \frac{-2 \pm \sqrt{4 - 4(1)(-2)}}{2} = \frac{-2 \pm \sqrt{12}}{2} = -1 \pm \sqrt{3}$.Checking the condition: $\sqrt{3} \approx 1.732$,so $x_1 = -1 + 1.732 = 0.732$ (not in $(-3, -1)$) and $x_2 = -1 - 1.732 = -2.732$ (in $(-3, -1)$).Thus,$x = -1 - \sqrt{3}$ is a solution.There are exactly $2$ real solutions.

The number of real solutions of the equation $|x^2 + 4x + 3| + 2x + 5 = 0$ is:

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