Question 982·Hard·Math

A solution to a system is an ordered pair that makes both equations true. So it appears at an intersection, where the two graphs meet. Count shared points, not points that lie on only one graph.

Type $x^2+y^2=10$ and $y=x^2-2$ on separate Desmos lines. Click the two crossings: Desmos labels them about $(-2.091,2.372)$ and $(2.091,2.372)$. Their $x$-coordinates differ, so these are two distinct real ordered pairs. The system has exactly two solutions. Choice B.

The second equation already gives $y$. Use substitution: put $x^2-2$ in place of $y$ in the first equation to keep both conditions:

Let $u=x^2$. A real number squared is never negative, so $u\ge0$. Replace both appearances of $x^2$:

Expand $(u-2)^2$, keeping its middle term:

Combine $u-4u$:

Subtract $10$ from both sides:

At $u=0$, the left side is $-6$, below zero. Its graph is an upward-opening parabola that rises on both ends, so it crosses the horizontal axis once on each side of $0$. Only the positive root can equal $x^2$.

Restore $x$ and $y$ from that one positive value of $u$:

The two signs give different $x$-coordinates because $u>0$, while $y$ is the same. So the system has exactly two distinct real ordered pairs. Choice B.