Question

If the roots of the eq 3x^2 + 2(k^2+1)x +(k^2-3k+2)=0 be opposite sign then the interval of k

Answer 1

by quadratic formula,$x=\frac {b^2\pm\sqrt{b^2-4ac}}{2a}$. hence if you want roots of opposite sign, then $b^2-4ac \ge b^2 => 4ac \le 0 => ac\le 0$.
therefore $k^2-3k+2\le 0 => (k-1)(k-2) \le 0 => 1\le k \le 2.$