SAT Mathematics Practice Test 2

Substituting 3 for y in $\mathit{y}=\mathit{a}{\mathit{x}}^2+\mathit{b}$ gives $3=\mathit{a}{\mathit{x}}^2+\mathit{b}$, which can be rewritten as $3-\mathit{b}=\mathit{a}{\mathit{x}}^2$.Since $\mathit{y}=3$ is one of the equations in the given system, any solution x of $3-\mathit{b}=\mathit{a}{\mathit{x}}^2$ corresponds to the solution $\left(\mathit{x},3\right)$ of the given system. Since the square of a real number is always non negative, and a positive number has two square roots, the equation $3-\mathit{b}=\mathit{a}{\mathit{x}}^2$ will have two solutions for x if and only if (1) $\mathit{a}>0$ and or (2) and $\mathit{b}>3$. Of the values for a and b given in the choices, only $\mathit{a}=-2,\mathit{b}=4$ satisfy one of these pairs of conditions.Alternatively, if $\mathit{a}=-2$ and $\mathit{b}=4$, then the second equation would be $\mathit{y}=-2{\mathit{x}}^2+4$.The graph of this quadratic equation in the xy-plane is a parabola with y-intercept $\left(0,4\right)$ that opens downward. The graph of the first equation, $\mathit{y}=3$, is the horizontal line that contains the point $\left(0,3\right)$. As shown below, these two graphs have two points of intersection, and therefore, this system of equations has exactly two real solutions. (Graphing shows that none of the other three choices produces a system with exactly two real solutions.)
Choices A, C, and D are incorrect and may result from calculation or conceptual errors.