`X_1;X_2` are the root of : `X^2-(2cost+3sint)X-11sin^2t=0` Find the minimum of `A=X_1^2+X_1X_2+X_2^2`.

`X_1;X_2` are the root of : `X^2-(2cost+3sint)X-11sin^2t=0` 

Find the minimum of `A=X_1^2+X_1X_2+X_2^2`.

Solution

We can see that: `A=X_1^2+X_1X_2+X_2^2`

                                 `=(X_1+X_2)^2-X_1X_2`

Following Vieta's Formulas `X^2-SX+P=0`    `(1)`

Then `X_1+X_2=S`  and `X_1X_2=P` which `X_1;X_2` are the roots of equation `(1)`

We easily see that: `X_1+X_2=2cost+3sint`

                                  `X_1X_2=-11sin^2t`

Therefore: `A=(2cost+3sint)^2+11sin^2t`

                     `=4cos^2t+12sintcost+11sin^2t+9sin^2t`

                     `=4cos^2t+12sintcost+20sin^2t`

                     `=4+6sin2t+16sin^2t`

                     `=4+6sin2t+8(1-cos2t)`

                     `=4+6sin2t+8-8cos2t`

                     `=12+2(3sin2t-4cos2t)`

                     `=12+10(3/5sin2t-4/5cos2t)`

                     `=12+10sin(2t-a)`   when `a=arcsin(4/5)`

As we knew: for all `t;a\inR` 

                    `-1\leqsinx``\leq1`  where is `x=2t-a`

Hence, Min`(A)=12-10=2`

Solution by: Thin Sokkean