Question 148·200 Super-Hard SAT Math Questions·Advanced Math

Try $k=146$ in Desmos. An intersection is a point on both graphs, so each crossing is a solution. Click the two crossings, $(71,11041)$ and $(75,11625)$. This value of $k$ gives two solutions, not none.

At any intersection, both expressions for $y$ give the same number, so set them equal:

Subtract the whole right side so the equation equals zero:

For $ax^2+bx+c=0$, the discriminant is $D=b^2-4ac$. Here $a=1$, $b=-k$, and $c=5325$. Substitute with the signs intact:

Simplify and require less than zero:

A quadratic has no real roots when its discriminant is negative, not zero; zero would give one touching point.

Add $21300$ to both sides:

Keep the strict inequality. A value at the boundary would still give one real solution.

Type $k_1^2\sim21300\{k_1>0\}$ in Desmos. The $\sim$ asks Desmos to find a value of $k_1$ that fits the equation; the restriction keeps the positive value. Under PARAMETERS, Desmos reports $k_1\approx145.945$. So $k$ must be below that boundary.

The greatest positive integer below $145.945$ is $145$. So $145$ is the greatest possible value of $k$ for which the line and parabola have no shared point. Choice B.