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The sum of all real roots of the equation $|x-3|^{2}+$ $|x-3|-2=0$ is
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6
$|x-3|^{2}+|x-3|-2=0$
Let $|x-3|=t$
$\therefore \mathrm{t}^{2}+\mathrm{t}-2=0 \Rightarrow \mathrm{t}^{2}+2 \mathrm{t}-\mathrm{t}-2=0$
$\Rightarrow \mathrm{t}(\mathrm{t}+2)-1(\mathrm{t}+2)=0$
$\Rightarrow(\mathrm{t}+2)(\mathrm{t}-1)=0$
$\Rightarrow \mathrm{t}=-2$ or $\mathrm{t}=1$
Since tis modulus of a number, it cannot be negative. $\therefore \mathrm{t}=1 \Rightarrow|\mathrm{x}-3|=1 \Rightarrow \mathrm{x}-3=1$ or $\mathrm{x}-3=-1$
$\Rightarrow \mathrm{x}=4$ or 2
Sum of roots $=4+2=6$
Let $|x-3|=t$
$\therefore \mathrm{t}^{2}+\mathrm{t}-2=0 \Rightarrow \mathrm{t}^{2}+2 \mathrm{t}-\mathrm{t}-2=0$
$\Rightarrow \mathrm{t}(\mathrm{t}+2)-1(\mathrm{t}+2)=0$
$\Rightarrow(\mathrm{t}+2)(\mathrm{t}-1)=0$
$\Rightarrow \mathrm{t}=-2$ or $\mathrm{t}=1$
Since tis modulus of a number, it cannot be negative. $\therefore \mathrm{t}=1 \Rightarrow|\mathrm{x}-3|=1 \Rightarrow \mathrm{x}-3=1$ or $\mathrm{x}-3=-1$
$\Rightarrow \mathrm{x}=4$ or 2
Sum of roots $=4+2=6$
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