Answer
We have a solution $c$ in $\left[\frac{1}{4}, 2\right]$
Work Step by Step
Let $f(x)=\sqrt{x}+\sqrt{x+2}-3 .$
Note that $f$ is continuous on $\left[\frac{1}{4}, 2\right]$ with $f\left(\frac{1}{4}\right)=\sqrt{\frac{1}{4}}+\sqrt{\frac{9}{4}}-3=-1$
and $f(2)=\sqrt{2}-1 \approx 0.41 .$
Therefore, by the IVT, there is a $c \in\left[\frac{1}{4}, 2\right]$ such that $f(c)=\sqrt{c}+\sqrt{c+2}-3=0 .$ Thus $\sqrt{c}+\sqrt{c+2}=3$ and hence the equation $\sqrt{x}+\sqrt{x+2}=3$ has a solution $c$ in $\left[\frac{1}{4}, 2\right]$.
