it's an even function ,i.e. f(alpha)=f(-alpha)
if alpha is a real zero, so is -alpha
there is only one real solution -----> then it must be x=0, substitue and we get k^2-1=0 , k= 1 or -1
these two values of k would give x^4 - x^2=0 and x^4 + 3x^2=0 respectively
the 1st eqn has 3 real...