Question 494·Hard·Algebra

Each equation represents a line. For no solution, the $x$ and $y$ coefficients must match in proportion, but the numbers on the right must not. Cross-multiply the coefficients to find when the lines could be parallel, without dividing by anything containing $k$:

Expand the left side to turn the condition into a quadratic equation:

Subtract $6$ from both sides: $k^2+k-8=0$. Its roots are the candidate values of $k$, but you still need to check that the lines don’t overlap.

The discriminant $b^2-4ac$ tells you how many real roots a quadratic has. Here $a=1$, $b=1$, and $c=-8$. Type $1^2-4(1)(-8)$ in Desmos; it prints $33$. Since $33>0$, the quadratic has two distinct real roots.

If the equations were the same line, the first equation’s $x$ coefficient and its constant would have to scale together from the second equation. Match those pairs:

Divide both sides by $4$:

Subtract $2$: $k=\frac{3}{2}$. This is the only value that could make the lines overlap.

Type $k=\frac{3}{2}$ and both original equations in Desmos. Click where the lines meet: Desmos shows $(2,0)$. At the only possible overlap value, the lines instead cross once. So neither real root of the parallel-line condition makes the equations identical; both give no solution.

For $ak^2+bk+c=0$, the sum of the roots is $-b/a$. In $k^2+k-8=0$, $a=1$ and $b=1$, so the two real values add to $-\frac{b}{a}=-\frac{1}{1}=-1$. The sum of all real values of $k$ that make the system have no solution is $-1$. Choice B.