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

Try squaring both sides of the equation to eliminate the square root, then rearrange the result into a standard quadratic equation in $x$. Remember to check any solutions you find back in the original equation.

Once you have a quadratic in $x$, use the discriminant $b^2 - 4ac$ to control how many real solutions it has. How should the discriminant relate to zero if the equation is to have exactly one real solution?

The discriminant will give you a condition on $n$. Use that to identify the smallest value of $n$ that allows at least one real solution, then verify that this value actually gives exactly one valid solution to the original equation and finally compute $8n$.

In Desmos, type a on a line by itself to create a slider (this will play the role of $n$). Then graph the two functions y = sqrt(a - 2x) and y = 40 - x. Make sure the square root graph appears only where a - 2x >= 0.

Move the slider for a left and right and watch how many intersection points the graphs have. You are looking for the smallest value of a for which the graphs intersect at least once.

Adjust a until the two graphs just touch at a single point (they are tangent). Note that value of a; that is the corresponding $n$ that gives exactly one real solution. Finally, multiply this value by 8 to match what the question asks for.

We are given

Because the square root is always nonnegative, the right-hand side must also be nonnegative:

• $40 - x \ge 0$, so $x \le 40$. And the expression inside the square root must be nonnegative:
• $n - 2x \ge 0$, so $x \le \dfrac{n}{2}$. Any solution must satisfy both of these inequalities and the equation itself.

Square both sides (this is valid as long as we later check for extraneous solutions):

Rearrange to get a quadratic in $x$:

So any real solution of the original equation must satisfy

A quadratic $ax^2 + bx + c = 0$ has real solutions when its discriminant $D = b^2 - 4ac$ is at least 0.

Here, $a = 1$, $b = -78$, and $c = 1600 - n$, so

For there to be any real solutions, we need $D \ge 0$:

So no values of $n$ less than 79 can give even one real solution to the original equation.

The smallest $n$ that allows real roots is $n = 79$. When $D = 0$, the quadratic has exactly one real root (a double root).

Plug $n = 79$ into the quadratic:

With $D=0$, the root is

Now check this $x$ in the original equation and the domain conditions:

• $40 - x = 40 - 39 = 1 \ge 0$ (okay),
• $n - 2x = 79 - 2\cdot39 = 79 - 78 = 1 \ge 0$ (okay),
• and

So $x = 39$ is a valid solution, and since the quadratic has only this one root, there is exactly one real solution for $n = 79$.

We have found that the smallest value of $n$ for which the equation has exactly one real solution is $n = 79$.

The problem asks for $8n$:

So the minimum possible value of $8n$ is 632.