Let's determine the real roots of the following equation with 3 significant digits:

$$
\sqrt{2 x-1}+\sqrt{x-4}=x-1
$$
(1) is defined for $x \geqq 4$. By trying some integer values, we find that $x=5$ is a root. To find further roots, the equation can be significantly simplified by introducing a new unknown in place of one of the square roots, as a linear expression stands under the roots, and thus $x$ will be a polynomial of the new unknown. It is advisable to introduce a new unknown in place of $\sqrt{x-4}$, and since its value at $x=5$ is 1, it is convenient to use $1+t$:

$$
\begin{gathered}
\sqrt{x-4}=1+t, \text { i.e., } \\
x=t^{2}+2 t+5, \text { where } t \geqq-1
\end{gathered}
$$

because the square root is non-negative. The equation will take the following form:

$$
\sqrt{2 t^{2}+4 t+9}=t^{2}+t+3
$$

From this, by squaring and rearranging, we arrive at the equation

$$
t\left(t^{3}+2 t^{2}+5 t+2\right)=0
$$

The value of $x$ corresponding to the root $t=0$ has already been found above, so the other roots must be sought from the following equation:

$$
f(t)=t^{3}+2 t^{2}+5 t+2=0
$$

taking into account the constraint $t \geqq-1$. For $t \geqq 0$, $f(t)>0$, so (4) has no positive roots. However, there is one in the interval $(-1,0)$, because $f(-1)=-20
\end{gathered}
$$

because all 6 terms in the square brackets are positive, so $f(t)$ is strictly increasing in the interval.

Based on the values $f(-1)=-2, f(0)=2$ and using linear interpolation, we try the approximation $t_{1}=-0.5$: $f(-0.5)=-0.125$, so the root is greater than $t_{1}$. However, taking $t_{2}=-0.4$, we get $f(-0.4)=0.256>0$, so the root is less than $t_{2}$. Using linear interpolation again based on the last two values, $t_{2}=-0.5+0.125 / 0.381 \approx-0.467$ is suggested as an approximation, and it turns out to be good: $f(-0.467)=0.00067$, so the root of (4) is $t>-0.467$.

From (2), $x>4.284 \ldots$, because according to the transformation $x=(t+1)^{2}+4$, $x$ is a monotonically increasing function of $t$ for $t>-1$.

To determine the value of $x$ to the second decimal place, consider the value of $f(-0.466 \ldots=f(-7 / 15)$: $2 / 3375>0$, so $t<-7 / 15$, and for $t=-7 / 15$, $x=3856 / 900<4.285$. Thus, to the desired accuracy, $x=4.28$, which indeed satisfies (1), and there are no further roots.

Remark. In (3),

$$
\sqrt{2 t^{2}+4 t+9}=y
$$

Rearranging this, the pairs of values satisfying the relation are represented by the points on the curve

$$
\frac{y^{2}}{7}-\frac{(t+1)^{2}}{7 / 2}=1
$$

in the $t, y$ coordinate system that satisfy the condition $y \geqq 0$. This is the upper branch of a hyperbola centered at $(-1,0)$ with a focal axis parallel to the $y$-axis.

The image of the right-hand side of (3) is the parabola

$$
y=t^{2}+t+3=\left(t+\frac{1}{2}^{2}\right)+\frac{11}{4}
$$

The two curves intersect at $t=0$, and between $-1 / 2$ and 0, as shown in the figure.