Question 344·Hard·Algebra

No Desmos needed. The sum-of-roots rule will give the requested sum after we check the roots. For parallel lines, one multiplier must match both the $x$ and $y$ coefficients. Cross-multiply the coefficient pairs from the given equations so you don’t have to divide by $k+1$: $(k+1)(k-4)=3(-2)=-6$. Matching coefficients finds parallel-line candidates; checking constants tells you whether the lines are distinct.

Expand the left side: $k^2-4k+k-4=-6$. Combine its $k$ terms: $k^2-3k-4=-6$.

Add $6$ to both sides to put the quadratic in a form you can factor: $k^2-3k+2=0$.

The numbers $-1$ and $-2$ multiply to $2$ and add to $-3$. Use them to factor: $(k-1)(k-2)=0$.

A zero product means at least one factor is zero: $k-1=0\quad\text{or}\quad k-2=0$. Solve each equation: $k=1\quad\text{or}\quad k=2$. These are candidates, not yet confirmed no-solution values.

At $k=1$, the equations are $2x-2y=4$ and $3x-3y=1$. Multiply the first equation by $\frac32$ to match the second equation’s left side: $3x-3y=6$. But the second equation says $3x-3y=1$. The constants disagree, so these parallel lines have no solution.

At $k=2$, the two equations become $3x-2y=4,\qquad 3x-2y=2$. Their left sides match, but their right sides disagree. So this value also gives distinct parallel lines with no solution.

Both candidates work. For $ak^2+bk+c=0$, the sum of the roots is $-\frac{b}{a}$. In $k^2-3k+2=0$, that sum is: $-\frac{-3}{1}=3$. The sum of all possible values of $k$ is $3$. Grid in 3.