Question

f(x) = 2x^(4) − 4x^(2) - 48. Find both of the real zeros of the polynomial function.

Answer 1

solve
\[2z^2-4z-48=0\] then replace
\[z=x^2\] and solve that one

Answer 2

I already did that but my answers don't make sense.

Answer 3

you get
\[2(z-6)(z+4)=0\]
\[z=6,z=-4\] so
\[x^2=6,x^2=-4\] the second one is not possible, so only two answers that make sense are
\[x^2=6\]
\[x=\pm\sqrt{6}\]