Question 196·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

A square root is never negative. It equals $40-x$, so any solution must satisfy $40-x\ge 0$:

Square both sides to get an equation without the radical:

Squaring can add false solutions, so we'll check the original equation later.

Add $2x$ to both sides so $n$ stands alone:

Expand the entire square. Its middle term is $-2(40)x=-80x$:

Combine the like terms, the two terms containing $x$:

Half of $-78$ is $-39$. To complete the square, use $(x-39)^2$ and subtract the extra $39^2$ it contains:

Type $1600-39^2$ in Desmos; it shows $79$, so:

A square cannot be less than $0$. So $n\ge 79$, with equality only when $(x-39)^2=0$, or $x=39$. That $x$ also meets the earlier limit $x\le 40$.

Type $n=79$ and graph both sides of the original equation in Desmos. Click where they meet: Desmos shows $(39,1)$. So $x=39$ works in the original, and the squared equation showed it is the only possible $x$ when $n=79$.

Type $8n$ on a new Desmos line; it prints $632$. Since $79$ is the smallest possible $n$ and gives exactly one solution, the minimum value of $8n$ is $632$. Grid in 632.