Estimating square roots

Learning outcome

Bracket nonnegative square roots, refine decimal estimates and justify rounding by comparing squared bounds and midpoints.

Concepts and assumptions

For N ≥ 0, √N denotes the nonnegative principal root. The radical is exact; a rounded decimal is approximate, so write ≈ instead of =. For N > 0, solving x² = N requires both signs, whereas a physical length uses the positive root.

Squaring preserves order for nonnegative numbers. If 0 ≤ a < b, then b² - a² = (b - a) × (b + a) > 0. Therefore a² < N < b² proves a < √N < b.

Begin with consecutive integer squares: n² ≤ N < (n + 1)². The integer square root is n, but n is not necessarily the nearest integer to √N. Refine by testing tenths, hundredths or smaller steps.

Rounding requires the requested precision. Between neighbouring candidates, square their midpoint to decide which side contains the true root. For an exact halfway case, round upward here. Comparing distances from N to two squares is not equivalent to comparing distances from √N to their roots.

Truncation simply removes later digits; rounding may increase the retained value. An answer rounded to the nearest 0.01 has an absolute error at most 0.005. The difference between its square and N is a different quantity, not the root error.

All inputs below are exact. Retain full precision in checks and round only the final root.

Worked examples

Example 1 — Nearest integer. Since 13² = 169 < 187 < 196 = 14², we have 13 < √187 < 14. The midpoint is 13.5, whose square is 182.25. Because 187 > 182.25, the root exceeds 13.5. Thus √187 ≈ 14 to the nearest whole number.

Example 2 — Nearest tenth. To estimate √58, calculate 7.6² = 57.76 and 7.7² = 59.29. Hence 7.6 < √58 < 7.7. Their midpoint is 7.65, and 7.65² = 58.5225 > 58. The root is below 7.65, so to the nearest 0.1, √58 ≈ 7.6.

Example 3 — A length with an error check. A square has area 123 m², so its positive side length is √123 m. Since 11² = 121 and 11.1² = 123.21, the side lies between 11 and 11.1 m. Refine: 11.09² = 122.9881 < 123, while 11.095² = 123.099025 > 123. Therefore 11.09 < √123 < 11.095. To the nearest 0.01 m, the side is approximately 11.09 m, with error less than 0.005 m.

Common mistakes

A lower bound is not automatically a rounded answer. Do not assume square roots change linearly. Report the requested number of decimal places and preserve the approximation sign.

Practice questions

1. Between which consecutive integers does √215 lie?
2. Round √73 to the nearest whole number.
3. Round √6 to the nearest 0.1 using a midpoint check.
4. Round √11 to the nearest 0.01 and justify an error less than 0.005.

Worked solutions

1. Since 14² = 196 < 215 < 225 = 15², we have 14 < √215 < 15.
2. The bounds 8² = 64 and 9² = 81 bracket 73. Since 8.5² = 72.25 < 73, √73 > 8.5. Thus √73 ≈ 9.
3. Squares 2.4² = 5.76 and 2.5² = 6.25 bracket 6. The midpoint square is 2.45² = 6.0025 > 6, so √6 < 2.45. Therefore √6 ≈ 2.4.
4. Since 3.31² = 10.9561 < 11 < 11.0224 = 3.32², test midpoint 3.315. Its square is 10.989225 < 11. Thus 3.315 < √11 < 3.32, giving √11 ≈ 3.32 with error less than 0.005.