The number of real solutions of the equation $| x^{2} + 4 x + 3 | + 2 x + 5 = 0$ are

1

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Solution

The correct option is D $2$
$f (x) = x^{2} + 4 x + 3 \Rightarrow x^{2} + 3 x + x + 3 \Rightarrow (x + 1) (x + 3) f (x) < 0 f o r x ϵ (- 3, - 1) | f (x) | + 2 x + 5 = 0 w h e n f (x) > 0 x^{2} + 6 x + 8 = 0 x^{2} + 4 x + 2 x + 8 = 0 (x + 4) (x + 2) = 0 x = - 4, - 2 b u t (f o r - 2, f (x) < 0) x = - 4 w h e n f (x) < 0 x ϵ (- 3 - 1) x^{2} + 4 x + 3 - 2 x - 5 = 0 x^{2} + 2 x - 2 = 0 x = - (1 + \sqrt{3})$
Two real solution.

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