Question 1120559
<br>
We need to have<br>
(9-0.62) < x^2 < (9+0.62)
sqrt(8.38) < x < sqrt(9.62)
2.8948 < x < 3.1016  (approximately)<br>
The upper limit is closer to 3 (3.1016-3 = 0.1016) than the lower limit (3-2.8948 = 0.1052).  So to guarantee that x^2 is within 0.62 of 9, x has to be within approximately 0.1052 of 3.