Certainly! Let's solve the problem of finding the set \( \{x : x \text{ is an integer and } x^2 < 40\} \).
We need to determine the integer values of \( x \) such that the square of \( x \) is less than 40.
1. Identify the range of \( x \) satisfying \( x^2 < 40 \):
- We want to find the integer values for which \( x^2 < 40 \).
- Start by estimating the maximum integer value for \( x \). Since \( x^2 < 40 \), \( x \) can be as large as the greatest integer less than \(\sqrt{40} \approx 6.32\). Thus, the largest integer \( x \) can be, while satisfying \( x^2 < 40 \), is 6.
- Remember to consider negative integers as well, because squaring negative numbers results in positive values. So the range must include negative integers.
2. Test integer values within the range \([-6, 6]\):
- The square of \(-6\) is \( (-6)^2 = 36 < 40 \).
- The square of \(-5\) is \( (-5)^2 = 25 < 40 \).
- The square of \(-4\) is \( (-4)^2 = 16 < 40 \).
- The square of \(-3\) is \( (-3)^2 = 9 < 40 \).
- The square of \(-2\) is \( (-2)^2 = 4 < 40 \).
- The square of \(-1\) is \( (-1)^2 = 1 < 40 \).
- The square of \( 0 \) is \( 0^2 = 0 < 40 \).
- The square of \( 1 \) is \( 1^2 = 1 < 40 \).
- The square of \( 2 \) is \( 2^2 = 4 < 40 \).
- The square of \( 3 \) is \( 3^2 = 9 < 40 \).
- The square of \( 4 \) is \( 4^2 = 16 < 40 \).
- The square of \( 5 \) is \( 5^2 = 25 < 40 \).
- The square of \( 6 \) is \( 6^2 = 36 < 40 \).
3. Conclusion:
- Hence, all integers \( x \) in the set \( \{ -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 \} \) satisfy \( x^2 < 40 \).
Therefore, the set of integers \( x \) such that \( x^2 < 40 \) is:
[tex]\[ \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\} \][/tex]
