8. It is known that the equation $2 x^{3}-7 x^{2}+7 x+p=0$ has three distinct roots, and these roots form a geometric progression. Find p and solve this equation.
Answer. $p=-2$; roots of the equation: $x=1, x=\frac{1}{2}, x=2$.

Solution Let $x, k x, k^{2} x$ be the roots of the given equation, with $x \neq 0, k \neq 0, k \neq \pm 1$ (since the roots are distinct). From Vieta's theorem, we have

$$
\left\{\begin{array}{l}
x+k x+k^{2} x=\frac{7}{2} \\
k x^{2}+k^{2} x^{2}+k^{3} x^{2}=\frac{7}{2} \\
k^{3} x^{3}=-\frac{p}{2}
\end{array}\right.
$$

Dividing the second equation of this system by the first, we get $k x=1$. Then the third equation gives $p=-2$. The original equation becomes $2 x^{3}-7 x^{2}+7 x-2=0$, from which $2(x-1)\left(x^{2}+x+1\right)-7 x(x-1)=0,(x-1)\left(2 x^{2}-5 x+2\right)=0$. Therefore, $x=1, x=\frac{1}{2}$ or $x=2$.

1 Plot the set of points on the coordinate plane whose coordinates satisfy the system of inequalities $\left\{\begin{array}{l}|| y y+5 x-15 \mid \leq 15, \\ |x|-3)^{2}+(|y|-5)^{2} \leq 1\end{array}\right.$ and find the area of the resulting figure.

Answer. $\frac{3 \pi}{2}$.

Solution The second inequality of the system defines four circles of radius 1 with centers at points $A(-3 ; 5)$, В $(3 ; 5), C(3 ;-5), D(-3 ;-5)$. The second inequality defines a strip between the parallel lines $y=-\frac{5 x}{3}$ and $y=10-\frac{5 x}{3}$. The line $y=-\frac{5 x}{3}$ passes through points $A$ and $C$, while the line $y=10-\frac{5 x}{3}$ passes through point $B$.

The set of points defined by the system is three semicircles of radius 1, so the area is $\frac{3 \pi}{2}$.