Question 188·Hard·Advanced Math

The given $xy=6$ and nonzero $x$ let you divide by $x$:

Use substitution: put the whole fraction $\frac{6}{x}$ wherever the second equation has $y$:

Multiply $3$ by $6$:

Multiply every term by $x$ to remove the fraction. This preserves the equation because the problem says $x\ne0$:

Subtract $kx$ from both sides:

In standard form $ax^2+bx+c=0$, the coefficients are $a=2$, $b=-k$, and $c=18$. Keep the minus sign on $b$.

The discriminant is $b^2-4ac$, the part under the quadratic formula's square root. Type $4(2)(18)$ in Desmos. It prints $144$, which must be subtracted from $b^2$.

For a two-root check, pick $(2,3)$ because $2\cdot3=6$. Type $k=2(2)+3(3)$, then graph both given equations in Desmos. It sets $k=13$; click the intersections to see $(2,3)$ and about $(4.5,1.333)$. Both satisfy the nonzero conditions, so keeping only one $x$ would miss a solution.

For $ax^2+bx+c=0$, the quadratic formula is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. The $\pm$ means add and subtract the square root. Substitute the coefficients and the $144$ you found:

Simplify the signs and the square:

Multiply the denominator:

These are the possible values of $x$ in terms of $k$ whenever the system has real solutions. Keep both roots unless a given condition rules one out. Choice C.