The number of real roots of the equation x|x – 2| + 3|x –3| + 1 = 0 [2025]

Question

The number of real roots of the equation $x$ $| x - 2 |$ $+ 3$ $| x - 3 |$ $+ 1 = 0$ [2025]

Smart Achievers · Accepted Answer

(4)

(I) When x < 2, we have

$- x^{2} + 2 x - 3 x + 9 + 1 = 0$

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

$\Rightarrow x = \frac{- 1 + \sqrt{41}}{2}, \frac{- 1 - \sqrt{41}}{2}; x = \frac{- 1 - \sqrt{41}}{2}$ ( $∵$ x < 2)

(II) When $2 \leq x < 3$ , we have

$x^{2} - 2 x - 3 x + 9 + 1 = 0$

$\Rightarrow x^{2} - 5 x + 10 = 0$

As D < 0 $\Rightarrow$ No real roots.

(III) When $x \geq 3$ , we have

$x^{2} - 2 x + 3 x - 9 + 1 = 0$

$\Rightarrow x^{2} + x - 8 = 0$

$\Rightarrow x = \frac{- 1 + \sqrt{33}}{2}, \frac{- 1 - \sqrt{33}}{2}$ (rejected)

Thus, only one real root exists.

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