Question 242·Hard·Advanced Math

Let $y$ stand for $k$ so Desmos can graph the system. Type $y-x=4$ and $x^2+xy=12$, then click both intersections, the points that satisfy both equations. Desmos shows about $(-3.646,0.354)$ and $(1.646,5.646)$. The second coordinate gives $k$, so keep both points.

The graph labels are rounded, so use the linear equation to find an exact sum. Add $x$ to both sides: $k=x+4$. Subtract $4$ from both sides:

Use substitution: each solution also satisfies $x^2+kx=12$, so replace both occurrences of $x$ with $k-4$:

Expand the square: $k^2-8k+16+k(k-4)=12$. Distribute $k$: $k^2-8k+16+k^2-4k=12$. Combine like terms:

Subtract $12$ from both sides: $2k^2-12k+4=0$. Divide every term by $2$: $k^2-6k+2=0$. These are the possible $k$-values, written as roots of a quadratic, an equation with a squared variable.

Call the two roots $k_1$ and $k_2$. Because the coefficient of $k^2$ is $1$, their factors expand as $(k-k_1)(k-k_2)=k^2-(k_1+k_2)k+k_1k_2$. Match the coefficient of $k$ to the $-6k$ in your equation: $-(k_1+k_2)=-6$. Multiply by $-1$: $k_1+k_2=6$. The middle coefficient gives the sum without finding either root separately. The sum of all possible values of $k$ is $6$. Grid in 6.