Estimating non-perfect square roots | Year 9

01 / LEARN

Understand the idea.

Start with nearby squares

Interpret $\sqrt{n}$ as the nonnegative side length of a square with area $n$ square units. Nearby perfect squares give an interval immediately: $25 < 30 < 36$ means the side lies between $5$ and $6$ units.

To narrow the interval, square decimal trial values. Do not divide the area gap proportionally: squaring is not linear, so halfway between two areas need not produce a side halfway between the two side lengths.

Check the rounding boundary

A one-decimal answer names the nearest tenth. The rounding boundaries for $5.5$ are $5.45$ and $5.55$ : values from the first boundary up to the second round to $5.5$ , using the usual rule for positive halfway values.

Compare the number under the root with the squares of these boundaries. This justifies the rounding without needing every decimal digit. Use $\approx$ for the rounded result; an inequality interval gives bounds rather than an exact decimal value.

a^{2} < n < b^{2} \Rightarrow a < \sqrt{n} < b (a, b \geq 0)

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A root between five and six

Estimate $\sqrt{30}$ to one decimal place.

Worked solution

1
Find the integer interval.
$5^{2} = 25 < 30 < 36 = 6^{2}$
Thus the nonnegative root is between 5 and 6.
2
Test the boundaries for rounding to 5.5.
$\begin{aligned} {5.45}^{2} & = 29.7025 \\ {5.55}^{2} & = 30.8025 \end{aligned}$
These are the squares of the two halfway boundaries around 5.5.
3
Locate the root within those boundaries.
$29.7025 < 30 < 30.8025 ⟹ 5.45 < \sqrt{30} < 5.55$
All values are positive, so the square comparisons preserve order.
4
Round to the nearest tenth.
$\sqrt{30} \approx 5.5$
Every value strictly between 5.45 and 5.55 rounds to 5.5.

Answer

$\sqrt{30} \approx 5.5$ .

EXAMPLE 02

Close to a perfect square

Estimate $\sqrt{80}$ to one decimal place.

Worked solution

1
Find the integer interval.
$8^{2} = 64 < 80 < 81 = 9^{2}$
The root is below 9 but above 8.
2
Test the boundaries for rounding to 8.9.
$\begin{aligned} {8.85}^{2} & = 78.3225 \\ {8.95}^{2} & = 80.1025 \end{aligned}$
The upper boundary decides whether the result rounds to 8.9 or 9.0.
3
Compare with 80.
$78.3225 < 80 < 80.1025 ⟹ 8.85 < \sqrt{80} < 8.95$
The positive root lies inside the interval that rounds to 8.9.
4
State the rounded result.
$\sqrt{80} \approx 8.9$
The root is below the halfway boundary 8.95.

Answer

$\sqrt{80} \approx 8.9$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Is $\sqrt{45}$ between $6.7$ and $6.8$ ?

Show answer & reasoning

Yes.

Yes: ${6.7}^{2} = 44.89 < 45 < 46.24 = {6.8}^{2}$ . Since all values are positive, this gives $6.7 < \sqrt{45} < 6.8$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Find consecutive whole-number bounds for $\sqrt{37}$ .
2 marks
Find consecutive whole-number bounds for $\sqrt{150}$ .
2 marks
Use squared trial values to locate $\sqrt{12}$ between tenths.
2 marks
Estimate $\sqrt{18}$ to one decimal place.
2 marks
Estimate $\sqrt{70}$ to one decimal place.
2 marks
A square has area $55 {cm}^{2}$ . Estimate its side to the nearest tenth of a centimetre.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find consecutive whole-number bounds for $\sqrt{37}$ .
2 marks
$36 < 37 < 49$ .
$6 < \sqrt{37} < 7$ .
Find consecutive whole-number bounds for $\sqrt{150}$ .
2 marks
$144 < 150 < 169$ .
$12 < \sqrt{150} < 13$ .
Use squared trial values to locate $\sqrt{12}$ between tenths.
2 marks
${3.4}^{2} = 11.56 < 12 < 12.25 = {3.5}^{2}$ .
$3.4 < \sqrt{12} < 3.5$ .
Estimate $\sqrt{18}$ to one decimal place.
2 marks
${4.15}^{2} = 17.2225 < 18 < 18.0625 = {4.25}^{2}$ .
$4.2$ .
Estimate $\sqrt{70}$ to one decimal place.
2 marks
${8.35}^{2} = 69.7225 < 70 < 71.4025 = {8.45}^{2}$ .
$8.4$ .
A square has area $55 {cm}^{2}$ . Estimate its side to the nearest tenth of a centimetre.
3 marks
The side is $\sqrt{55} cm$ .
${7.35}^{2} = 54.0225 < 55 < 55.5025 = {7.45}^{2}$ .
$7.4 cm$ .

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Year 9 real-number classification and applications; root bounds and sensible digital approximation. Full surd manipulation is developed in Year 10.

Official curriculum source ↗

In this lesson

Understand the idea.

Start with nearby squares

Check the rounding boundary

See the method in action.

A root between five and six

Find the integer interval.

Test the boundaries for rounding to 5.5.

Locate the root within those boundaries.

Round to the nearest tenth.

Close to a perfect square

Find the integer interval.

Test the boundaries for rounding to 8.9.

Compare with 80.

State the rounded result.

Try it yourself.

Check your understanding

Keep practising