Question 84·200 Super-Hard SAT Math Questions·Advanced Math

For each fixed $a$, a solution is an $x$ that makes the equation true. In Desmos, type $\sqrt{y-x}=30-2x$, using $y$ to stand for $a$. Desmos draws the allowed $(x,a)$ pairs as a curve with a low point. A horizontal height through one point on that curve represents one solution.

The square root is nonnegative, so $30-2x\ge0$, or $x\le15$. Square both sides: $a-x=(30-2x)^2$. Add $x$: $a=x+(30-2x)^2$. Keep the original restriction after squaring. Otherwise, values with a negative right side could be counted as solutions.

Expand the whole square, including its middle term: $(30-2x)^2=900-120x+4x^2$. Substitute it into the expression for $a$: $a=x+900-120x+4x^2$. Combine the $x$ terms: $a=4x^2-119x+900$. This quadratic opens upward because its $x^2$ coefficient is positive, so it has a lowest point.

Type $y=4x^2-119x+900\{x\le15\}$; the braces tell Desmos to show only allowed $x$-values. Click the vertex, the curve’s lowest point. Desmos labels it about $(14.875,14.938)$. Its $x$-value is within $x\le15$, so this low point belongs to the original equation.

A quadratic $Ax^2+Bx+C$ reaches its vertex at $x=-B/(2A)$. Here $A=4$ and $B=-119$, so substitute: $x=\frac{-(-119)}{2(4)}$. Simplify: $x=\frac{119}{8}=14.875$. This matches the graph’s input. Use the exact fraction, not the graph’s rounded height, to find $a$.

At this lowest height, the horizontal line touches the allowed curve once; below it, the line misses the curve. Put $x=119/8$ into $a=x+(30-2x)^2$, and type $16\left(\frac{119}{8}+\left(30-2\cdot\frac{119}{8}\right)^2\right)$ in Desmos. It prints $239$. So the minimum possible value of $16a$ is $239$. Choice B.