The parabolas defined by the equations $y=2x^2-4x+4$ and $y=-x^2-2x+4$ intersect at points $(a,b)$ and $(c,d)$, where $c\ge a$. What is $c-a$? Express your answer as a common fraction.
To find the points of intersection, I need to set the two parabolas equal to each other and solve for $x$. So I have $2x^2-4x+4=-x^2-2x+4$, which simplifies to $3x^2-2x=0$. I can factor this as $x(3x-2)=0$, so $x=0$ or $x=\frac{2}{3}$. These are the values of $a$ and $c$, and since $c\ge a$, I know that $c=\frac{2}{3}$ and $a=0$. Therefore, $c-a=\frac{2}{3}-0=\frac{2}{3}$. 