Question 120·200 Super-Hard SAT Math Questions·Algebra

Test $k=-4$ by typing $k=-4$, $y=\left|2x-k\right|$, and $y=x+4$ in Desmos. The graphs have two intersections. Each intersection gives an $x$ where the original sides are equal; you need a value of $k$ that leaves one.

The vertex is the tip of the V, where the expression inside the bars is zero. Since $2\left(\frac{k}{2}\right)-k=0$, the tip is at $\left(\frac{k}{2},0\right)$.

Moving one unit away from the tip raises the V by $2$, while the line's height changes by only $1$. So a line above the tip meets both arms, and one below it meets neither. Exactly one intersection occurs when the line passes through the tip. The tip's height is $0$, so set the line's height there to $0$:

Type $\frac{k_1}{2}+4\sim0$. Here $k_1$ is the unknown you're finding, and $\sim$ tells Desmos to fit it. Desmos shows $k_1=-8$ under PARAMETERS. That value puts the line through the V's tip, giving exactly one real solution for $x$. Choice B.