Question 126·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

An intersection is a point on both graphs, so the two expressions for $y$ must be equal there:

Subtract all of $mx+1$ from both sides:

The $+1$ moves with $mx$.

Combine the constants and both terms containing $x$:

The $-mx$ belongs in the $x$ coefficient; don't leave it out.

For $ax^2+bx+c=0$, the discriminant is $b^2-4ac$, the part under the quadratic formula's square root. One intersection means one $x$-root, so the discriminant is $0$. Here $a=1$, $b=-(m+6)$, and $c=12$. Set $b^2=4ac$:

Multiply $4(1)(12)$:

Take both square roots, since a positive and a negative number can have the same square:

Since $48=16\cdot3$ and $\sqrt{16}=4$, pull $4$ out of the square root:

Subtract $6$ from each possible value of $m+6$:

To check $m>0$, type $(m_1+6)^2\sim48\{m_1>0\}$ in Desmos. The $\sim$ asks Desmos to fit $m_1$, and the braces restrict it to positive values. Under PARAMETERS, it shows $m_1\approx0.928203$. Type $4\sqrt3-6$ on the next line; Desmos prints the same decimal. The minus branch is negative, so the required slope is $4\sqrt3-6$. Choice D.