The roots of $x^{2}+b x+c=0$ are the squares of the roots of $x^{2}-5 x+2=0$. What is the value of $\frac{c}{b}$ ?
For the quadratic equation $x^{2}-d x+e=0$, the sum of the roots is $d$ and the product of the roots is $e$.

This is because if the roots are $x=f$ and $x=g$, then

$$
x^{2}-d x+e=(x-f)(x-g)=x^{2}-f x-g x+f g=x^{2}-(f+g) x+f g
$$

Suppose that $r$ and $s$ are the roots of $x^{2}-5 x+2=0$.

Then $r+s=5$ and $r s=2$.

We are told that the roots of $x^{2}+b x+c=0$ are $r^{2}$ and $s^{2}$.

Then

$$
c=r^{2} \cdot s^{2}=(r s)^{2}=2^{2}=4
$$

and

$$
-b=r^{2}+s^{2}=(r+s)^{2}-2 r s=5^{2}-2(2)=21
$$

which tells us that $b=-21$.

Therefore, $\frac{c}{b}=\frac{4}{-21}=-\frac{4}{21}$.

(We could also solve the problem by directly calculating the roots of $x^{2}-5 x+2=0$, squaring these roots, and creating a quadratic equation with these results as roots.)

ANSWER: $-\frac{4}{21}$