Single Variable Calculus, Chapter 2, 2.5, Section 2.5, Problem 54

(a) Show that the equation $\displaystyle \sqrt{x - 5} = \frac{1}{x + 3}$ has at least one real root.
(b) Determine the root using a graph.

(a) Let $f(x) = \displaystyle \sqrt{x - 5} - \frac{1}{x + 3} $
Based from the definition of Intermediate value Theorem,
There exist a solution $c$ for the function between the interval $(a,b)$ suppose that the function is continuous
on the given interval. So we take $a$ and $b$ to be 5 and 6 respectively and assume the function $f(x)$
is continuous on the interval (5,6). So we have,


$
\begin{equation}
\begin{aligned}

f(5) =& \sqrt{5 - 5} - \frac{1}{5 + 3} = \frac{-1}{8} < 0\\
& \text{ and }\\
f(6) =& \sqrt{6 - 5} - \frac{1}{6 + 3} = \frac{8}{9} > 0

\end{aligned}
\end{equation}
$


By using Intermediate Value Theorem. We prove that...

So,
$
\begin{equation}
\begin{aligned}
& \text{if } 5 < c < 6 && \text{then } \quad f(5) < f(c) < f(6)\\
& \text{if } 5 < c < 6 && \text{then } \quad -\frac{1}{8} < 0 < \frac{8}{9}
\end{aligned}
\end{equation}
$

Therefore,
There exist such root for $\displaystyle \sqrt{x - 5} = \frac{1}{x + 3}$.

(b) Referring to the graph, we can estimate the root of the function as $x = 5.03$.