Mathematics - Quadratic Equation Question with Solution | TestHub

For one root to be greater than 2 and the other less than 2, the value of the quadratic function $f(x)=x^2-2(k-1)x+k+5$ at $x=2$ must be negative (since the leading coefficient is positive). So, $f(2)<0$. Substituting $x=2$: $2^2-2(k-1)(2)+k+5<0$. This simplifies to $4-4k+4+k+5<0$, which means $-3k+13<0$. Therefore, $3k>13$, or $k>\frac{13}{3}$.