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

y  =  2x^2 - 4x + 4    and    y  =  -x^2 - 2x + 4

 

First let's find the  x  values of the intersection points by solving this equation:

 

2x^2 - 4x + 4   =   -x^2 - 2x + 4

                                                  Add  x^2  to both sides of the equation.

3x^2 - 4x + 4   =   -2x + 4

                                                  Add  2x  to both sides.

3x^2 - 2x + 4   =   4

                                                  Subtract  4  from both sides.

3x^2 - 2x  =  0

                                                  Factor  x  out of the two terms on the left side.

x(3x - 2)   =   0

                                                  Set each factor equal to zero.

x  =  0     or     3x - 2  =  0

                       3x  =  2

                         x  =  2/3

 

The  x  coordinates of the intersection points are  0  and  2/3 .

So   c = 2/3   and   a = 0

 

c - a   =   2/3 - 0   =   2/3

 

Here's a graph to check it:  https://www.desmos.com/calculator/enndawdskv