# 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.

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