Question 129·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Type $x^2+y^2=10$ and $y=x^2-4$ on separate Desmos lines. Desmos draws a circle and a parabola, a U-shaped graph. Keep the entire circle visible so you can check where either half meets the parabola.

Click every point where the graphs cross. Desmos labels them approximately $(-2.449,2)$, $(-1,-3)$, $(1,-3)$, and $(2.449,2)$. Each distinct intersection is one ordered pair that satisfies both equations. So four distinct real ordered pairs satisfy the system. Choice D.

The second equation gives $y=x^2-4$, so substitute that whole expression for $y$ in the first equation: $x^2+(x^2-4)^2=10$. Keep the parentheses because the entire expression for $y$ is squared.

Let $u=x^2$ to make the repeated $x^2$ easier to work with. Replace each $x^2$:

Expand the square: $u+u^2-8u+16=10$. Combine the $u$ terms: $u^2-7u+16=10$. Subtract $10$: $u^2-7u+6=0$. This is a quadratic, an equation with a squared variable, in $u$ rather than $x$.

Factor using $1$ and $6$, whose product is $6$ and whose sum is $7$: $(u-1)(u-6)=0$. By the zero-product property, a product is zero if at least one factor is zero, so keep both cases: $u-1=0\quad\text{or}\quad u-6=0$. Add $1$ in the first case and $6$ in the second:

You defined $u=x^2$, so put $x^2$ back in both cases: $x^2=1\quad\text{or}\quad x^2=6$. These are equations for $x$, not yet the ordered pairs.

Take both signs of each square root, since a positive and a negative number have the same square: $x=\pm1\quad\text{or}\quad x=\pm\sqrt6$. Keeping only positive roots would lose valid points.

For $x^2=1$, use $y=x^2-4$: $y=1-4=-3$. For $x^2=6$, use the same equation: $y=6-4=2$. So the pairs are $(-1,-3)$, $(1,-3)$, $(-\sqrt6,2)$, and $(\sqrt6,2)$. They are four distinct ordered pairs satisfying the system. Choice D.