Electrical – How to find the intersection point in rootlocus diagram

I'm not sure about your calculation but Matlab yields the correct result. In this problem, the common approach is to use Routh-Hurwitz criterion and search for a row of zeros that yields the possibility for imaginary axis roots. For convert the system to the closed-loop transfer function, hence $$ \frac{K}{s^4 + 10s^3 + 88s^2 + 256s + K} $$

The Routh table is

$$ \begin{matrix} s^4 &&&& 1 &&&& 88 &&&& K \\ s^3 &&&& 10 &&&& 256 \\ s^2 &&&& 62.4 &&&& K \\ s^1 &&&& \frac{15974.4-10K}{62.4} \\ s^0 &&&& K \end{matrix} $$

The \$ s^1 \$ row is the only row that can yield a row of zeros. From the preceding row, we obtain

$$ \begin{align} &15974.4 - 10K = 0 \\ K &= \frac{15974.4}{10} = 1597.44 \end{align} $$

Now we take a look at the row above \$s^1\$ and construct the following polynomial, hence

$$ \begin{align} 62.4 s^2 + K &= 0 \\ 62.4 s^2 + 1597.44 &= 0 \\ s^2 &= \frac{-1597.44}{62.4} \\ s_{1,2} &= \pm j \sqrt{25.6} \\ s_{1,2} &= \pm j 5.0596 \\ \end{align} $$

The root locus crosses the imaginary axis at \$\pm j5.0596\$ at the gain \$K=1597.44\$. Consequently, the gain \$K\$ must be less than 1597.44 for the system to be stable.

No - the root locus is not drawn for the open-loop function.

It is the main purpose of the root locus of a system with feedback to show if resp. for which constant gain values the CLOSED-LOOP function will have poles in the "forbidden" region (right half of the s-plane). Hence, you have to find the poles of the function

H(s)=G(s)/[1+G(s)*H(s)] with G=1/(s+3)(s+4) and H=K(s(s+1).

That means, you have to find the zeroes of the denominator which gives the characteristic equation: 1+G(s)*H(s)=0