\(X_1,X_2\) are the roots of \(X^2-(2\cos t+3\sin t)X-11\sin^2t=0\). Find the minimum value of \(A=X_1^2+X_1X_2+X_2^2\).

Solution

\(X_1,X_2\) are the roots of \(X^2-(2\cos t+3\sin t)X-11\sin^2t=0\). By Vieta's formulas, we have \(X_1+X_2=2\cos t+3\sin t\) and \(X_1X_2=-11\sin^2t\).

We have \(A=X_1^2+X_1X_2+X_2^2=(X_1+X_2)^2-X_1X_2\).

\[ \begin{aligned} A&=(2\cos t+3\sin t)^2+11\sin^2t\\ &=4\cos^2t+12\sin t\cos t+9\sin^2t+11\sin^2t\\ &=4\cos^2t+12\sin t\cos t+20\sin^2t\\ &=4+6\sin2t+16\sin^2t\\ &=4+6\sin2t+8(1-\cos2t)\\ &=12+2(3\sin2t-4\cos2t)\\ &=12+10\left(\frac35\sin2t-\frac45\cos2t\right). \end{aligned} \]

Let \(\sin\alpha=\frac45,\ \cos\alpha=\frac35\). Then

\[ \frac35\sin2t-\frac45\cos2t=\sin(2t-\alpha). \]

Therefore,

\[ A=12+10\sin(2t-\alpha). \]

Since \(-1\leq\sin(2t-\alpha)\leq1\), we obtain

\[ 2\leq A\leq22. \]

Hence, the minimum value of \(A\) is

\[ \boxed{\min A=2}. \]

Solution by: Thin Sokkean