The quadratic $x^2 + 5x + c$ has roots in the form of $x = \frac{-5 \pm \sqrt{c}}{2}$. What is the value of $c$?
Using the quadratic formula $x = \frac{ - b \pm \sqrt {b^2 - 4ac} }{2a}$, we can find the roots of the quadratic. We find that $x = \frac{-5 \pm \sqrt{25-4c}}{2}$. Thus, setting our two expressions for $x$ equal to each other, we find that  \begin{align*}
\frac{-5 \pm \sqrt{25-4c}}{2} &= \frac{-5 \pm \sqrt{c}}{2} \quad \Rightarrow \\
25 - 4c &= c \quad \Rightarrow \\
c &= \boxed{5}.
\end{align*}
5
Level 4
Algebra
Problem: The quadratic $x^2 + 5x + c$ has roots in the form of $x = \frac{-5 \pm \sqrt{c}}{2}$. What is the value of $c$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of \( c \), we start with the quadratic equation \( x^2 + 5x + c \). The roots of a quadratic equation in the standard form \( ax^2 + bx + c = 0 \) can be found using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 5 \), and \( c \) is the constant we need to determine.

Substituting the values into the quadratic formula gives us:

\[
x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot c}}{2 \cdot 1}
\]

This simplifies to:

\[
x = \frac{-5 \pm \sqrt{25 - 4c}}{2}
\]

We are given that the roots are in the form \( x = \frac{-5 \pm \sqrt{c}}{2} \). For these two expressions for the roots to be equivalent, the terms under the square root must be equal. Therefore, we set the expressions under the square root equal to each other:

\[
\sqrt{25 - 4c} = \sqrt{c}
\]

Next, we square both sides to eliminate the square roots:

\[
25 - 4c = c
\]

Now, we solve for \( c \) by rearranging the equation:

\[
25 = c + 4c
\]
\[
25 = 5c
\]
\[
c = \frac{25}{5}
\]
\[
c = 5
\]

Thus, the value of \( c \) that satisfies the condition of the roots is \( 5 \).<|END_OF_LATENT|> The final answer is $5$.