, 2023

The quadratic function $f(x) = x^2 + (2n-1)x + (n^2 - 22)$ has no real roots exactly when its
discriminant, $\Delta$, is
negative.

The discriminant of this function is $$\begin{align*} \Delta & = (2n-1)^2 - 4(1)(n^2 - 22) \\ & = (4n^2 - 4n + 1) - (4n^2 - 88) \\ & = -4n + 89\end{align*}$$ We have $\text{lt; 0}$ exactly when $\text{-4n + 89 lt; 0}$ or $\text{4n gt; 89}$.

This final inequality is equivalent to $\text{n gt; 89 4 = 22 1 4}$.

Therefore, the smallest positive integer that satisfies this inequality,
and hence for which $f(x)$ has no
real roots, is $n = 23$.
Using the cosine law in $\triangle PQR$, $$\begin{align*} PR^2 & = PQ^2 + QR^2 - 2 \cdot PQ \cdot QR \cdot \cos(\angle PQR) \\ 21^2 & = a^2 + b^2 - 2ab\cos(60^\circ) \\ 441 & = a^2 + b^2 - 2ab\cdot \tfrac{1}{2} \\ 441 & = a^2 + b^2 - ab\end{align*}$$ Using the sine law in
$\triangle STU$, we obtain $\dfrac{ST}{\sin(\angle TUS)} = \dfrac{TU}{\sin(\angle TSU)}$ and so $\dfrac{a}{4/5} = \dfrac{b}{\sin(30^\circ)}$.

Therefore, $\dfrac{a}{4/5} = \dfrac{b}{1/2}$ and so $a = \tfrac{4}{5} \cdot 2b = \tfrac{8}{5}b$.

Substituting into the previous equation, $$\begin{align*} 441 & = \left(\tfrac{8}{5}b\right)^2 + b^2 - \left(\tfrac{8}{5}b\right)b \\ 441 & = \tfrac{64}{25}b^2 + b^2 - \tfrac{8}{5}b^2 \\ 441 & = \tfrac{64}{25}b^2 + \tfrac{25}{25}b^2 - \tfrac{40}{25}b^2 \\ 441 & = \tfrac{49}{25}b^2 \\ 225 & = b^2\end{align*}$$ Since $\text{b gt; 0}$, then $b = 15$ and so $a = \tfrac{8}{5}b = \tfrac{8}{5} \cdot 15 = 24$.