The number of real roots of the equation x ∣ x − 2 ∣ + 3 ∣ x − 3 ∣ + 1 = 0 is:

Question:

The number of real roots of the equation

$$x\mathrm{\mid }x-2\mathrm{\mid }+3\mathrm{\mid }x-3\mathrm{\mid }+1=\mathrm{0\; is:}$$

📷 Question Image:

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Short Solution (Text):

Critical points for the absolute values are $x=2$ and $x=3$. Solve piecewise in the three intervals.

1) Region $x<2$:
$|x-2|=2-x,\; |x-3|=3-x$. The equation becomes

$$x(2-x) + 3(3-x) + 1 = 0 \\ \Rightarrow -x^2 - x + 10 = 0 \\ \Rightarrow x^2 + x - 10 = 0.$$

Discriminant $\Delta = 1 + 40 = 41$. Roots:

$$x = \frac{-1 \pm \sqrt{41}}{2}.$$

Numeric values: $\dfrac{-1+\sqrt{41}}{2}\approx 2.7016$ and $\dfrac{-1-\sqrt{41}}{2}\approx -3.7016$.
Only $x\approx -3.7016$ lies in this region ($x<2$) → one valid root here.

2) Region $2 \le x < 3$:
$|x-2|=x-2,\; |x-3|=3-x$. The equation becomes

$$x(x-2) + 3(3-x) + 1 = 0 \\ \Rightarrow x^2 - 5x + 10 = 0.$$

Discriminant $\Delta = 25 - 40 = -15 <0$ → no real roots in this interval.

3) Region $x\ge 3$:
$|x-2|=x-2,\; |x-3|=x-3$. The equation becomes

$$x(x-2) + 3(x-3) + 1 = 0 \\ \Rightarrow x^2 + x - 8 = 0.$$

Discriminant $\Delta = 1 + 32 = 33$. Roots:

$$x = \frac{-1 \pm \sqrt{33}}{2}\approx 2.3723,\; -3.3723.$$

Neither root satisfies $x\ge 3$ → no valid roots here.

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✅ Final Answer:
Number of real roots = 1
(the only real solution is $x = \dfrac{-1-\sqrt{41}}{2}\approx -3.7016$).