#[sqrt(70)]
6 messages · Page 1 of 1 (latest)
√64 < √70 < √81
8 < √70 < 9
√70 = 8 + (some decimal)
[√70] = 8
If you want a general method to approximate square roots:
√x ≈ [√x] + (x - [√x]²)/2[√x]
Works best when x is slightly larger than a perfect square, gets more accurate for large x