The parabolas defined by the equations $y=x^2+4x+6$ and \(y=\frac{1}{2}x^2+x+6\) intersect at points $(a,b)$ and $(c,d)$, where \(c\ge a\). What is c-a?

Set the functions equal

(1/2)x^2 + x + 6 = x^2 + 4x + 6 rearrange as

(1/2)x^2 + 3x = 0

Factor

x [ (1/2)x + 3 ] = 0

Setting both factors equal to 0 and solving for x, we have that

x = 0 and x = -6

c = 0 and a = - 6

So c - a = 0 - -6 = 6