The diagonals of a parallelogram bisect each other, then AE=CE. Since [tex]AE = x^2-8 [/tex] and [tex]CE = 2x [/tex], you have that
[tex]x^2-8=2x[/tex].
Solve this equation:
[tex]x^2-2x-8=0, \\ D=(-2)^2-4\cdot (-8)=4+32=36, \\ \sqrt{D}=6, \\ x_{1,2}=\dfrac{2\pm 6}{2} =-2,4[/tex].
If x=-2, then CE=-4 that is impossible? because CE is distance and should be positive.
If x=4, then CE=8 and AE=8. The whole diagonal AC=AE+CE=8+8=16.
