It's a Plastic

Let $a = \sqrt[3]{\frac 12 + \sqrt{\frac jk}}$ and $b = \sqrt[3]{\frac 12 - \sqrt{\frac jk}}$ . Then $x=a+b$ and

$\begin{aligned} x^3 & = x + 1 \\ (a+b)^3 & = a+b+1 \\ {\color{#3D99F6}a^3} + 3a^2b + 3ab^2 + {\color{#3D99F6}b^3} & = a+b + 1 & \small \color{#3D99F6} \text{Note that }a^3+b^3 = 1 \\ 3ab(a+b) + {\color{#3D99F6}1} & = a+b + 1 \\ 3ab(a+b) & = a+b \\ ab & = \frac 13 \\ \sqrt[3]{\frac 12 + \sqrt{\frac jk}}\times \sqrt[3]{\frac 12 - \sqrt{\frac jk}} & = \frac 13 \\ \sqrt[3]{\frac 14 - \frac jk} & = \frac 13 & \small \color{#3D99F6} \text{Cubing both sides} \\ \frac 14 - \frac jk & = \frac 1{27} \\ \frac jk & = \frac 14 - \frac 1{27} \\ & = \frac {23}{108} \end{aligned}$

Therefore, $j = 23$ , $k=108$ and $\sqrt{16 + jk} = \sqrt {16+23\times 108} = \sqrt{2500} = \boxed{50}$ .

We can solve the equation from the outset, but given that the value of $x$ is partially given, we can utilize it by cube it, which gives $x^3 = (\sqrt[3]{\frac{1}{2}+\sqrt{\frac{j}{k}}}+\sqrt[3]{\frac{1}{2}-\sqrt{\frac{j}{k}}})^3 = 1+3x(\sqrt[3]{\frac{1}{4}-(\sqrt{\frac{j}{k}})^2})$ By comparing the coefficient to the original equation, we will get that $3(\sqrt[3]{\frac{1}{4}-\frac{j}{k}}) = 1$ or simply $\frac{1}{4}-\frac{j}{k} = \frac{1}{27}$ , which can be reduced further to $\frac{j}{k} = \frac{23}{108}$ . Given that both integers are relative prime, it is then concluded that $(j,k) = (23,108)$ and $\sqrt{16+jk} = \sqrt{16+2484} = \sqrt{2500} = 50$

Note: If one try to resolve the equation from the outset without using $x$ provided partially as a guideline, using the Remainder Factor Theorem and Complex Conjugates will result into that two factor of this equation must be the complex number and another as a real number without rationalization, so one may want to set $x = a+b$ or $x^3 = a^3+3ab(x)+b^3 = x+1$ as a rule. Here, it is easy to observe that $ab = \frac{1}{3}$ and $a^3+b^3 = 1$ or $a^3+\frac{1}{27a^3} = 1$ , which may be reduced into $27a^6-27a^3+1 = 0 = 27(a^6-a^3+\frac{1}{4}-\frac{1}{4})+1$ Or, when trying to put it into perfect square and reducing the coefficient $(a^3-\frac{1}{2})^2-\frac{23}{108}=0$ Which means $a = \sqrt[3]{\frac{1}{2}+\sqrt{\frac{23}{108}}};\:b = \sqrt[3]{\frac{1}{2}-\sqrt{\frac{23}{108}}}$ Notice that if you swap $a$ and $b$ , the result will be the same. The rest is followed by the above solution. It is worth nothing that this number is called Plastic number , hence the name of the question.