Find the set of all values of $a$ for which the equation $x^2-2ax+a^2+a-3=0$ has both roots less than $3$.

Solution

1. For the quadratic $f(x)=x^2-2ax+a^2+a-3$ to have two real roots less than 3, we require the discriminant $D\ge 0$, the vertex $x_v<3$, and $f(3)>0$.
2. The discriminant $$\begin{array}{rl}D& =(-2a{)}^2-4(a^2+a-3)\\ & =12-4a\ge 0\end{array}$$ implies $a\le 3$.
3. The vertex $$\begin{array}{rl}{x}_v& =-(-2a)/2\\ & =a<3\end{array}$$ is satisfied for $a<3$.
4. Evaluating $$\begin{array}{rl}f(3)& =9-6a+a^2+a-3\\ & =a^2-5a+6>0\end{array}$$ factors to $(a-2)(a-3)>0$, which implies $a<2$ or $a>3$.
5. Intersecting the conditions $a\le 3$, $a<3$, and ($a<2$ or $a>3$) results in $a<2$.
6. Hence the answer is (A).