# Multiple Scenarios

## If $x$ is a real number such that $x^2 > 10$, which of the following statements must be true?

A)$x > \sqrt{10}$

B)$x^2 + 2x > 10$

C)$x^2 + 2x > 3$

D)$x = 4$

E)$x^2 + 2x \neq 5$

If you tried to do this question quickly, you likely got tricked and chose the wrong answer. Options A and B appear a very tempting choice, but they do not satisfy all possible scenarios.

Let's see how to approach this problem. If $x^2 > 10$, then we know that either $x > \sqrt{10}$ or $x < - \sqrt{10}$.Consider statement A. $x = -4$ satisfies $x^2 > 10$. However, it does not satisfy $x > \sqrt{10}$.

Consider statement B. $x = -4$ satisfies $x^2 > 10$. However, it does not satisfy $x^2 + 2x > 10$.

Consider statement D. $x = -4$ satisfies $x^2 > 10$. However, it does not satisfy $x = 4$.

Consider statement E. $x = -1 - \sqrt{6}$ satisfies $x^2 > 10$. However, it does not satisfy $x^2 + 2x \neq 5$.

From here, we get that only Statement C could be true, so that is our answer. We now proceed to show it. If $x > \sqrt{10}$, then clearly $x^2 + 2x > x^2 = 10 > 3$.

If $x < - \sqrt{10}$, then $x+1 < - \sqrt{10} + 1 < 0$ and so $(x+1)^2 > ( -\sqrt{10} + 1)^2$. This gives us $x^2 + 2x + 1 > 10 - 2 \sqrt{10} + 1$. Hence, we get that $x^2 + 2x > 10 - 2 \sqrt{10} > 3$.Thus, the answer is C.

**Cite as:**Multiple Scenarios.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/multiple-scenarios/