# Counter Examples

The Multiple Scenarios skill is very similar to Case checking and Counter Examples, though it may not be immediately obvious how to proceed. You will have to carefully determine the scenarios, and consider each of them before making the final judgement.

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. $_\square$

**Cite as:**Counter Examples.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/counter-examples/