Example 5 (6th "Hope Cup" Invitational Competition Question) If the equation $x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=a$ has exactly one real root with respect to $x$, then the range of the real number $a$ is ().
A. $\left[\frac{1}{4},+\infty\right)$
B. $\left[\frac{1}{2},+\infty\right)$
C. $[1,+\infty)$
D. $[2,+\infty)$
Solution: Choose A. Reason: Since $\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=\frac{\sqrt{4 x+2+2 \sqrt{4 x+1}}}{2}=$ $\frac{1}{2}(\sqrt{4 x+1}+1)^{2}$, the original equation can be transformed into $2 x+\sqrt{4 x+1}+1=2 a$, which simplifies to $\sqrt{4 x+1}=-2 x+2 a-1$.
By examining the function $y_{1}=\sqrt{4 x+1}$, we know that $x \geqslant-\frac{1}{4}, y_{1} \geqslant 0$.
By examining the function $y_{2}=-2 x+2 a-1$ (or by plotting the graphs of $y_{1}$ and $y_{2}$), we can see that the original equation has one real root (i.e., the graphs of $y_{1}$ and $y_{2}$ intersect at only one point) if and only if $2 a-1 \geqslant 2 x=-\frac{1}{2}$, solving which gives $a \geqslant \frac{1}{4}$.