## Algebra 2 (1st Edition)

$$\pm1$$
For cross multiplication, we know: $$\frac{a}{d} =\frac{b}{c} \rightarrow ac=bd$$ Thus, we find: $$xx=-\left(x^2-2\right)\cdot \:1 \\ x^2=-\left(x^2-2\right)\\ x=\pm1$$ Checking our solution by plugging the given value back into the equation, we assure that the solutions are not extraneous.