Why are the solutions of the $3\times 3$ system like that? Consider the problem:$$
\min \quad -x_1^2-4x_1x_2-x_2^2\\ \text{s.t.} \quad x_1^2 + x_2^2 = 1$$
The KKT system is given by\begin{align*}
x_1 (-1 + v) + 2 x_2 &= 0 \tag{1} ,\\ x_2 (-1 + v) + 2 x_1 &= 0 \tag{2},\\ x_1^2 + x_2^2 &= 1 \tag{3}
\end{align*}
The solutions according to the book are$$
(x_1,x_2)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right),\ v=3;\\
(x_1,x_2)=\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right),\ v=3;\\
(x_1,x_2)=\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right),\ v=-1,\\
(x_1,x_2)=\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right),\ v=-1,\\
$$
I tried to solved by hand: Substracting (1) and (2) I get $(x_1-x_2)(1+v)=0$. As $v$ cannot be zero (by the KKT condition) we have that $x_1=x_2$ if $v\neq -1$.
Thus I get the solutions:$$
(x_1,x_2)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right);\\
(x_1,x_2)=\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right);\\
(x_1,x_2)=\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right);\\
(x_1,x_2)=\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right);\\
$$
only asking $v\neq -1$.
Why the book ask $v=-1$ and $v=3$? Could someone explain please?

Using polar coordinates,
$$\begin{aligned} x_1 &= \cos (\theta)\\ x_2 &= \sin (\theta)\end{aligned}$$
we obtain the unconstrained $1$-dimensional maximization problem in $\theta$
$$\text{maximize} \quad 1 + 2 \sin (2\theta)$$
Differentiating the objective and finding where the derivative vanishes, we obtain $\cos(2\theta) = 0$, whose solution set is
$$\left\{ \frac{\pi}{4} + k \frac{\pi}{2} : k \in \{0,1,2,3\} \right\}$$
Thus, in terms of $x_1$ and $x_2$, the solution set is
$$\left\{ \pm \left(\frac{\sqrt 2}{2},\frac{\sqrt 2}{2}\right), \pm \left(\frac{\sqrt 2}{2},-\frac{\sqrt 2}{2}\right)\right\}$$
which is the same you obtained via other means.
