$1^{\circ}$. Show that the roots of the equation

$$
x^{2}+(\lambda-2) x-(\lambda+3)=0
$$

are real for any (real) value of $\lambda$.

$2^{\circ}$. Express the sum of the squares of the roots as a function of $\lambda$!

$3^{\circ}$. Determine the value of $\lambda$ so that the sum of the squares of the roots is equal to a given number $k$! What is the smallest possible value of $k$?
$1^{0}$. The discriminant of the equation

$$
(\lambda-2)^{2}+4(\lambda+3)=\lambda^{2}+16
$$

is positive for any real value of $\lambda$, so the roots are real and distinct.

$2^{0}$. The sum of the squares of the roots

$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(2-\lambda)^{2}-2(\lambda+3)=\lambda^{2}-2 \lambda+10=(\lambda-1)^{2}+9
$$

$3^{0}$.

$$
(\lambda-1)^{2}+9=k, \quad \text { if } \lambda=1 \pm \sqrt{k-9} .
$$

$\lambda$ is real if $k \geqq 9$. Therefore, the smallest value of $k$ is 9, and in this case $\lambda=1$.