Solution
= Solution
Replace the quadratic constraint by the two linear inequalities $-2\le x_1-x_2+x_3\le2$. Simplex pivots lead to the active constraints $x_1=0$, $-x_2+x_3=2$, and $3x_2+x_3=4$. Thus $x^*=(0,1/2,5/2)$ and the value is $3$. With right-hand side $4+\varepsilon$, the same <basis> remains optimal for small nonzero $\varepsilon$, giving $x^*=(0,1/2+\varepsilon/4,5/2+\varepsilon/4)$.