Nonlinear equations and systems — SAT practice questions

Rewriting gives $x^3-4x=0$, so $x(x-2)(x+2)=0$. The three distinct real solutions are $-2$, 0, and 2.

Equating the expressions for $y$ gives $3x^2+42x-4-k=0$. Exactly one intersection means the discriminant is 0: $1764-12(-4-k)=0$, so $1812+12k=0$ and $k=-151$.

Squaring both sides gives $x^2+3x=0$, so the candidates are 0 and $-3$. Substituting $x=-3$ into the original equation gives $1 \neq -1$, so $x=-3$ is extraneous. The solution is $x=0$.