Question 132·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

At a solution, both equations give the same $y$ for the same $x$. So set their right-hand sides equal: $x^2+kx+4=-x^2+6x+1$. Each real $x$ that works here gives one point on both graphs.

Subtract the right-hand side from both sides, keeping the minus sign outside the parentheses: $x^2+kx+4-(-x^2+6x+1)=0$. Combine like terms:

A quadratic $ax^2+bx+c=0$ has exactly one real root when its discriminant, $b^2-4ac$, is $0$. Here $a=2$, $b=k-6$, and $c=3$. Substitute: $(k-6)^2-4(2)(3)=0$. Simplify: $(k-6)^2-24=0$. Add $24$ to both sides:

Graph $y=(x-6)^2$ and $y=24$. On this graph, the horizontal coordinate stands for $k$. Click both intersections: Desmos shows about $(1.101,24)$ and $(10.899,24)$. Those are two possible constants, not two solutions to the original system for one constant. The graph's coordinates are rounded, so keep the algebra for the exact sum.

Taking square roots of $(k-6)^2=24$ requires both signs: $k-6=\pm\sqrt{24}$. Add $6$ to both sides: $k=6\pm\sqrt{24}$. So the two values sit the same distance from $6$, on opposite sides.

The question asks for the sum, not either value alone. The opposite square roots cancel: $(6-\sqrt{24})+(6+\sqrt{24})=2(6)$. Type $2(6)$ on a new Desmos line; it shows $12$. When two values are equally far from a center, their sum is twice that center. The sum of all such values of $k$ is $12$. Grid in 12.

Starting from $(k-6)^2=24$, multiply out the square: $k^2-12k+36=24$. Subtract $24$ from both sides: $k^2-12k+12=0$. Its two roots are the two allowable values of $k$.

For $k^2+Bk+C=0$, the sum of the roots is $-B$. Here the coefficient of $k$ is $-12$, so the sum is $-(-12)=12$. That's the sum of all allowable values of $k$. Grid in 12.