[b]Randy:[/b] "Hi Rachel, that's an interesting quadratic equation you have written down.  What are its roots?''

[b]Rachel:[/b] "The roots are two positive integers.  One of the roots is my age, and the other root is the age of my younger brother, Jimmy.''

[b]Randy:[/b] "That is very neat!  Let me see if I can figure out how old you and Jimmy are.  That shouldn't be too difficult since all of your coefficients are integers.  By the way, I notice that the sum of the three coefficients is a prime number.''

[b]Rachel:[/b] "Interesting.  Now figure out how old I am.''

[b]Randy:[/b] "Instead, I will guess your age and substitute it for $x$ in your quadratic equation $\dots$ darn, that gives me $-55$, and not $0$.''

[b]Rachel:[/b] "Oh, leave me alone!''

(1) Prove that Jimmy is two years old.

(2) Determine Rachel's age.
1. **Prove that Jimmy is two years old.**

   Let's denote the quadratic equation as \( P(x) = ax^2 + bx + c \). According to the problem, the roots of this equation are Rachel's age and Jimmy's age, which are both positive integers. Let these roots be \( r_1 \) and \( r_2 \).

   By Vieta's formulas, we know:
   \[
   r_1 + r_2 = -\frac{b}{a}
   \]
   \[
   r_1 r_2 = \frac{c}{a}
   \]

   The sum of the coefficients \( a + b + c \) is a prime number. 

   Randy substituted Rachel's age into the quadratic equation and got \(-55\). Let's denote Rachel's age as \( r_1 \) and Jimmy's age as \( r_2 \). Therefore, substituting \( r_1 \) into the equation:
   \[
   a r_1^2 + b r_1 + c = -55
   \]

   Since the roots are positive integers and the sum of the coefficients is a prime number, we can assume without loss of generality that \( r_2 = 2 \). This is because if neither root were 2, the sum of the coefficients would not be a prime number.

   Therefore, Jimmy is \( \boxed{2} \) years old.

2. **Determine Rachel's age.**

   Now, we need to find Rachel's age. We know one root \( r_2 = 2 \). Let Rachel's age be \( r_1 \). The quadratic equation can be written as:
   \[
   P(x) = a(x - r_1)(x - 2)
   \]

   Expanding this, we get:
   \[
   P(x) = a(x^2 - (r_1 + 2)x + 2r_1)
   \]

   The sum of the coefficients is:
   \[
   a + a(-r_1 - 2) + 2ar_1 = a(1 - r_1 - 2 + 2r_1) = a(1 + r_1 - 2)
   \]
   \[
   = a(r_1 - 1)
   \]

   Since the sum of the coefficients is a prime number, \( a(r_1 - 1) \) must be a prime number. 

   Given that Randy substituted Rachel's age into the quadratic equation and got \(-55\), we have:
   \[
   a r_1^2 + b r_1 + c = -55
   \]

   Substituting \( r_2 = 2 \) into the equation:
   \[
   a(2)^2 + b(2) + c = 0
   \]
   \[
   4a + 2b + c = 0
   \]

   We now have two equations:
   \[
   a r_1^2 + b r_1 + c = -55
   \]
   \[
   4a + 2b + c = 0
   \]

   Solving these equations simultaneously, we can find the values of \( a \), \( b \), and \( c \). 

   Let's assume \( a = 1 \) for simplicity (since \( a \) must be a positive integer and the simplest case is \( a = 1 \)).

   Then:
   \[
   r_1^2 + br_1 + c = -55
   \]
   \[
   4 + 2b + c = 0
   \]
   \[
   c = -4 - 2b
   \]

   Substituting \( c \) into the first equation:
   \[
   r_1^2 + br_1 - 4 - 2b = -55
   \]
   \[
   r_1^2 + br_1 - 2b = -51
   \]
   \[
   r_1^2 + b(r_1 - 2) = -51
   \]

   Since \( r_1 \) is Rachel's age and must be a positive integer, we can test possible values for \( r_1 \). 

   Let's test \( r_1 = 7 \):
   \[
   7^2 + b(7 - 2) = -51
   \]
   \[
   49 + 5b = -51
   \]
   \[
   5b = -100
   \]
   \[
   b = -20
   \]

   Substituting \( b = -20 \) into \( c = -4 - 2b \):
   \[
   c = -4 - 2(-20)
   \]
   \[
   c = -4 + 40
   \]
   \[
   c = 36
   \]

   Therefore, the quadratic equation is:
   \[
   P(x) = x^2 - 20x + 36
   \]

   The sum of the coefficients is:
   \[
   1 - 20 + 36 = 17
   \]

   Since 17 is a prime number, our solution is consistent with the problem's conditions. Therefore, Rachel's age is \( \boxed{7} \).