L6 · Estimating square and cube roots

Learning goal

“I can estimate the square root or cube root of a number that is not a perfect power — to the nearest whole number and to the nearest tenth — by bracketing it between perfect squares or cubes and refining with test values.”

Benchmark — a known perfect square or cube used as a reference point: for √150 the benchmarks are 144 and 169.

Approximation — a value close to the exact value, written with ≈ ("is approximately"): √150 ≈ 12.2. An approximation is not the number itself.

Consecutive integers — whole numbers that follow one another with nothing between: 12 and 13 are consecutive; every non-perfect square root lives between two of them.

Worked example

Estimate √150 to the nearest tenth. Step 1: Bracket with consecutive squares: 12² = 144 < 150 < 169 = 13², so 12 < √150 < 13. Step 2: Position sense: 150 is 6 above 144 and 19 below 169 — the root hugs the bottom of its bracket, so expect ≈ 12.2, not 12.5. Step 3: Refine with tenths: 12.2² = 148.84 (distance from 150: 1.16); 12.3² = 151.29 (distance: 1.29). The smaller distance wins: √150 ≈ 12.2. Say it plainly: the estimate is a claim, and squaring is the audit. Every "≈" in this lesson is backed by two squared bounds.

Check yourself — explain it

150 is not halfway between 144 and 169, and the answer is not 12.5. What actually decides between 12.2 and 12.3?

Paired problem — example still visible

Estimate √62 to the nearest tenth. -> ≈ 7.9 (7² = 49 < 62 < 64 = 8². Tenths: 7.8² = 60.84, distance 1.16; 7.9² = 62.41, distance 0.41 -> 7.9.)

Faded completion — last step is yours

"√5: 2² = 4 < 5 < 9 = 3², so √5 is between ___ and ___. 2.2² = 4.84 (distance ___) and 2.3² = 5.29 (distance ___), so √5 ≈ ___." -> 2; 3; 0.16; 0.29; 2.2

Ready to practise?

10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.

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Grade 10 — Mathematics 10C › C10U04 Factors, Roots & Irrational Numbers

L6 · Estimating square and cube roots