Mathematics - Quadratic Equation Question with Solution | TestHub

For the quadratic equation $x^2-(k-3)x+k=0$ to have real and distinct roots, the discriminant $D>0$. $D=(-(k-3){)}^2-4(1)(k)=(k-3{)}^2-4k=k^2-6k+9-4k=k^2-10k+9$. So, $k^2-10k+9>0\Rightarrow (k-1)(k-9)>0$, which implies $k\in (-\infty ,1)\cup (9,\infty )$. For both roots to be positive, two conditions must be met: Sum of roots $>0$ and Product of roots $>0$. Sum of roots: $-(k-3)/1=k-3>0\Rightarrow k>3$. Product of roots: $k/1=k>0$. Combining all conditions: $k\in ((-\infty ,1)\cup (9,\infty ))$ AND $k>3$ AND $k>0$. The intersection of these conditions is $k\in (9,\infty )$.