The number of distinct real roots of the equation | x + 1 | | x + 3 | - 4 | x + 2 | + 5 = 0 is ________. [2024]

Question

The number of distinct real roots of the equation $| x + 1 | | x + 3 | - 4 | x + 2 | + 5 = 0$ is ________. [2024]

Smart Achievers · Accepted Answer

(2)

$| x + 1 | | x + 3 | - 4 | x + 2 | + 5 = 0$

$(I) If x < - 3, x^{2} + 4 x + 3 + 4 x + 8 + 5 = 0$

$\Rightarrow x^{2} + 8 x + 16 = 0 \Rightarrow x = - 4 (one solution)$

$(II) If - 3 \leq x < - 2, - x^{2} - 4 x - 3 + 4 x + 8 + 5 = 0$

$\Rightarrow x^{2} - 10 = 0 \Rightarrow x = \pm \sqrt{10}$

$which do not satisfy - 3 \leq x < - 2$

$(III) If - 2 \leq x < - 1, - x^{2} - 4 x - 3 - 4 x - 8 + 5 = 0$

$\Rightarrow x^{2} + 8 x + 6 = 0 \Rightarrow {(x + 4)}^{2} = 10$

$\Rightarrow x = - 4 \pm \sqrt{10}$ which do not satisfy $- 2 \leq x < - 1$

$(IV) If x \geq - 1, x^{2} + 4 x + 3 - 4 x - 8 + 5 = 0$

$\Rightarrow x^{2} = 0 \Rightarrow x = 0 (one solution)$

$Hence, the number of distinct real roots are two.$

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