Assume the roots of the quadratic are \( x_1 \) and \( x_2 \), then:
\[
x_1 + x_2 = -b, \quad x_1 x_2 = c.
\]
From the quadratic \( x^2 + 3x + 2 = 0 \), we get:
\[
x_1 + x_2 = -3, \quad x_1 x_2 = 2.
\]
Using identities:
\[
x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2 = 9 - 4 = 5.
\]
Also,
\[
x_1^3 + x_2^3 = (x_1 + x_2)^3 - 3x_1 x_2(x_1 + x_2) = (-3)^3 - 3(2)(-3) = -27 + 18 = -9,
\]
\[
x_1^5 + x_2^5 = (x_1 + x_2)(x_1^4 + x_2^4) - x_1 x_2(x_1^3 + x_2^3),
\]
which gives:
\[
3(-27) = 11(-9) \Rightarrow -81 = -81.
\]
Hence the quadratic \( x^2 + 3x + 2 = 0 \) satisfies all conditions.