Let $b$ and $c$ be real numbers.  If the polynomial $x^2+bx+c$ has exactly one real root and $b=c+1$, find the value of the product of all possible values of $c$.
Consider the quadratic formula $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Since the quadratic has exactly one root, then its discriminant must be 0. Thus, this gives us  \begin{align*} 0&=b^2-4ac
\\\Rightarrow\qquad0&=(c+1)^2-4c
\\\Rightarrow\qquad0&=(c^2+2c+1)-4c
\\\Rightarrow\qquad0&=c^2-2c+1
\\\Rightarrow\qquad0&=(c-1)^2.
\end{align*}Since this expression is a perfect square, the only possible of value of $c$ is 1. Thus, the product of all possible values of $c$ is $\boxed{1}$.
1
Level 4
Algebra
Problem: Let $b$ and $c$ be real numbers.  If the polynomial $x^2+bx+c$ has exactly one real root and $b=c+1$, find the value of the product of all possible values of $c$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the conditions under which the polynomial $x^2 + bx + c$ has exactly one real root, we need to analyze the discriminant of the polynomial. The discriminant $\Delta$ is given by the formula $\Delta = b^2 - 4c$. For the polynomial to have exactly one real root, the discriminant must equal zero: 

1. Set the discriminant to zero: 
   $b^2 - 4c = 0$.

2. Substitute the condition $b = c + 1$ into the discriminant equation:
   $(c + 1)^2 - 4c = 0$.

3. Expand the left-hand side:
   $c^2 + 2c + 1 - 4c = 0$.

4. Simplify the equation:
   $c^2 - 2c + 1 = 0$.

5. Factor the quadratic:
   $(c - 1)^2 = 0$.

6. Solve for $c$:
   $c - 1 = 0 \implies c = 1$.

Since the problem asks for the product of all possible values of $c$, and we have found that the only value of $c$ that satisfies the conditions is $1$, the product of all possible values of $c$ is simply $1$.<|END_OF_LATENT|> The final answer is $1$.