Question 168·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

At a solution, both equations give the same $y$, so their right sides are equal:

Subtract $x^2+5x$ from both sides to isolate $k$:

Combine the $x$ terms:

For each $x$, the line $y=x+7$ fixes one $y$. So exactly one solution point requires exactly one $x$ that produces the chosen $k$. The expression $7-x^2-4x$ makes a parabola that opens downward because its $x^2$ term is negative. A horizontal level $k$ reaches it only once when it touches its highest point. Below that point, the level crosses twice.

Type $y=7-x^2-4x$ and click the vertex, the turning point. Desmos shows $(-2,11)$. The second coordinate is the highest value of the expression for $k$, so $k=11$ makes the original system meet at exactly one point. Choice C.

At a shared point, $x^2+5x+k=x+7$. Subtract $x+7$ to put the equation in standard form, with $0$ on one side: $x^2+4x+k-7=0$. Each root, or $x$ that makes this equation true, gives one shared point because the line fixes its $y$.

For $ax^2+bx+c=0$, the discriminant $b^2-4ac$ is zero when the quadratic has exactly one real root. Here $a=1$, $b=4$, and $c=k-7$, so the condition is:

Type $4^2-4(1)(k-7)\sim0$ in Desmos. The $\sim$ tells Desmos to find the parameter that makes the two sides match. Under PARAMETERS, it shows $k=11$, so the value giving one solution is $11$. Choice C.