IB maths question - please help

It looks like these curves intersect at (0,0) and (k,k).
So integrate y=(1/k) x^2 between x=0 and x=k,
Integrate y=(kx)^ 1/2 between x=0 and x=k
Subtract the two areas which are now in terms of k and equate to 12.
Whence k.

LATER EDIT
Better still, the enclosed area is symmetrically bisected by the line y=x
Therefore Integrate y=(1/k) ^2,between x=0 and x=k,
Subtract the area under the line (triangle area 1/2 k^2)
Double the result (symmetetry) and equate to 12
Whence k.

LATER EDIT (typos)
Better still, the enclosed area is symmetrically bisected by the line y=x
Therefore Integrate y=(1/k)x ^2,between x=0 and x=k,
Subtract FROM the area under the line (triangle area 1/2 k^2)
Double the result (symmetry) and equate to 12
Whence k

Nice observation about the symmetry. Definitely simplifies the problem

Nice! Thanks so much for your time - this question was really bothering me. My main problem was that I had not calculated the intersects, meaning I was just left with a load of algebra as my answer.
Cheers

IB maths question - please help

i feel your pain, math irritates my bowels too