, 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 $\Delta < 0$ exactly when $-4n + 89 < 0$ or $4n > 89$. This final inequality is equivalent to $n
> $\frac{89}{4} = 22\frac{1}{4}$. 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$,$$\text{PR 2 = PQ 2 + QR 2 - 2 PQ QR ( PQR) 21 2 = a 2 + b 2 - 2ab (60 ) 441 = a 2 + b 2 - 2ab 1 2 441 = a 2 + b 2 - ab}$ 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 $b > 0$, then $b = 15$ and so $a = \tfrac{8}{5}b = \tfrac{8}{5} \cdot 15 = 24$.