How can Sign be factored?

The absoulte value fuction can be rewritten as: $|x|= x\text{sgn}(x)$

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Therefore:

$|x^2-3x+2| = (x^2-3x+2)\text{sgn}(x^2-3x+2)$

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Plugging this into our equation:

$\text{sgn}(x^2-3x+2)-(x^2-3x+2)\text{sgn}(x^2-3x+2)=0$

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Factoring:

$\text{sgn}(x^2-3x+2)(1-(x^2-3x+2))=0$

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The signum function is equal to zero whenever its argument is equal to zero, so for that we just have to solve:

$x^2-3x+2 = 0$

$x = 1, 2$

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Then the other part is already a simple quadratic:

$(1-(x^2-3x+2))=0$

$x^2-3x+1 = 0$

$\frac{3}{2} \pm \frac{\sqrt{5}}{2}$

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Therefore:

$1 + 2 + (\frac{3}{2} + \frac{\sqrt{5}}{2}) + (\frac{3}{2} - \frac{\sqrt{5}}{2}) = \fbox{6}$

Given that

$\text{sgn}(x^2 - 3x+2) - \mid x^2 - 3x + 2 \mid = 0 \implies \mid x^2 - 3x + 2 \mid = \text{sgn} (x^2-3x+2) = \begin{cases} -1 \\ 0 \\ 1 \end{cases}$

When $\text{sgn} (x^2-3x+2) = - 1$ , $\implies \mid x^2 - 3x + 2 \mid = -1$ . Since $\mid x^2 - 3x + 2 \mid \ge 0$ . there is no solution.

When $\text{sgn} (x^2-3x+2) = 0$ ,

$\begin{aligned} \mid x^2 - 3x + 2 \mid & = 0 \\ x^2 - 3x + 2 & = 0 \\ (x-1)(x-2) & = 0 \\ \implies x & = \begin{cases} 1 \\ 2 \end{cases} \end{aligned}$

When $\text{sgn} (x^2-3x+2) = 1$ ,

$\mid x^2 - 3x + 2 \mid = 1 \implies x^2 - 3x + 2 = \pm 1 \implies \begin{cases} x^2 - 3x + 1 = 0 & \implies x = \dfrac {3\pm \sqrt 5}2 \\ x^2 - 3x + 3 = 0 & \red{\text{No real roots.}} \end{cases}$

Therefore the sum of real roots is $1 + 2 + 3 = \boxed 6$ .

$sgn(x^2-3x+2) = |x^2-3x+2|$

Consider 3 cases $x^2-3x+2 = 0, 1, -1$

Rearrange $x^2-3x + c = 0$ where $c = 1, 2, 3$

$\Delta = (-3)^2 - 4 \times 1 \times c = 9 -4c$ which is $< 0$ when $c = 3$ .

sum of roots for each case $= - (-3) / 1 = 3.$

The question requires the sum of all real roots $= 3 + 3 = \boxed{6}$ for $c= 1, 2$