8th Grade · 8.NS.A.2

Estimating square roots in 8th grade

Estimating square roots means finding a close decimal for a root such as $\sqrt{30}$ , whose decimal never ends. The square root of 30 is the number that gives 30 when multiplied by itself, and no whole number or fraction does that exactly. In 8th grade, your child narrows a root down to a decimal and uses it to compare numbers and place them on a number line.

What the standard says

The official Common Core wording, quoted exactly. The plain version follows.

8.NS.A.2 Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions (e.g., π²).

What it means, in plain words

Estimating square roots starts with the perfect squares: 1, 4, 9, 16, 25, 36, 49 and so on. A number that falls between two perfect squares has a root that falls between their roots. For example, 40 is between 36 and 49, so $\sqrt{40}$ is between 6 and 7.

Approximating irrational numbers then goes one decimal place at a time. Take $\sqrt{2}$ , which is between 1 and 2. Squaring decimals closes the gap: $1.4^2 = 1.96$ and $1.5^2 = 2.25$ , so $\sqrt{2}$ is between 1.4 and 1.5. Next, $1.41^2 = 1.9881$ and $1.42^2 = 2.0164$ , so it is between 1.41 and 1.42. Each round gives one more digit, and the digits never end, because $\sqrt{2}$ is irrational.

These estimates let children work with irrational numbers. They can place irrational numbers on a number line, compare two of them and estimate the value of an expression such as $\pi^2$ . An estimate only settles a comparison when the two ranges do not overlap. If both numbers are between 3 and 4, the child needs another decimal place.

Your child should be able to:

Find the two whole numbers a square root lies between, such as 9 and 10 for $\sqrt{90}$
Narrow an estimate to tenths and then hundredths by squaring decimals
Place numbers such as $\sqrt{3}$ and $\pi$ on a number line
Compare two irrational numbers and estimate an expression such as $\pi^2$

Example

Estimate $\sqrt{30}$ to the nearest tenth.

30 is between the perfect squares 25 and 36, so $\sqrt{30}$ is between 5 and 6.
$5.4^2 = 29.16$ and $5.5^2 = 30.25$ , so $\sqrt{30}$ is between 5.4 and 5.5.
To see which is closer, square the middle: $5.45^2 = 29.7025$ , which is less than 30.
So $\sqrt{30}$ is more than 5.45, and to the nearest tenth it is about 5.5.

Example

Which is larger, $\sqrt{10}$ or $\pi$ ?

Both are between 3 and 4, so whole numbers cannot decide.
$3.1^2 = 9.61$ and $3.2^2 = 10.24$ , so $\sqrt{10}$ is between 3.1 and 3.2. $\pi = 3.14...$ is in the same range.
Go one place further: $3.16^2 = 9.9856$ and $3.17^2 = 10.0489$ , so $\sqrt{10}$ is between 3.16 and 3.17.
$\pi$ is about 3.14, so $\sqrt{10}$ is the larger number.

Where children usually get stuck

Your child says $\sqrt{30}$ is about 15.
The square root gets confused with half of the number. But 15 squared is 225, nowhere near 30.
What helps. Write the perfect squares from 1 to 144 on a card and keep it on the table. Before any estimate, your child finds the two squares on either side of the number.
For $5 \times \sqrt{2}$ , your child writes $5 + 1.41 = 6.41$ .
The estimate for $\sqrt{2}$ is right, but the operation changed on the way. Once the root becomes a decimal, the child stops looking at the original expression.
What helps. Ask your child to copy the expression with the decimal in place of the root, keeping every sign: $5 \times 1.41 = 7.05$ .
Your child finds that $\sqrt{14}$ and $\sqrt{15}$ are both between 3 and 4 and decides they are equal.
The child takes "in the same range" to mean "the same number". Whole-number bounds are too wide to tell these two apart.
What helps. Ask your child to go one decimal place further, or to compare the numbers under the root. 15 is more than 14, so $\sqrt{15}$ is the larger root.

Try four practice questions

Real 8th grade questions from our question bank, easiest first. Answer them right here, check each one, and see a worked solution for any question.

1Warm-up
Using only the given bounds, which statement correctly explains the student's claim?
$\displaystyle 1.41 < \sqrt{2} < 1.42 \qquad \text{Student claim: } 4.23 < 3\sqrt{2} < 1.42$
A
The claim is not guaranteed; one original endpoint was left unchanged.
B
The claim is not guaranteed; its endpoints match adding 3 instead of multiplying by 3.
C
The claim is not guaranteed; both endpoints lie strictly inside the transformed interval.
D
The claim is not guaranteed because a wider interval is never valid.
E
The claim is not guaranteed; the correctly calculated endpoints are in reverse order.
F
The claim is correct; it contains the whole interval found from the bounds.
Show the answer and how to solve it
The upper bound is still $1.42$ , the original endpoint. It must also be multiplied by $3$ , giving $4.26$ . The claim is not guaranteed.
1. 1Use the given bounds for $\sqrt{2}$ to check the claim about $3\sqrt{2}$ .
2. 2Multiply all three parts by $3$ . The multiplier is positive, so keep the inequality directions.
3. 3The upper bound is still $1.42$ , the original endpoint. It must also be multiplied by $3$ , giving $4.26$ . The claim is not guaranteed.
Common mistake. Judge what the given bounds guarantee. Do not assume unshown work or use the exact decimal value of the irrational.
The idea underneath. Transform both bounds, then check whether the claim contains the whole resulting interval. A wider interval can still be guaranteed.
2On level
Match each square root to the pair of consecutive integers that contains its value. One pair on the right will not be used.
Connect each item on the left to its match on the right.
Square roots
Integer pairs
$\displaystyle \sqrt{28}$
$\displaystyle \sqrt{67}$
$\displaystyle \sqrt{150}$
$\displaystyle 8 \text{ and } 9$
$\displaystyle 12 \text{ and } 13$
$\displaystyle 9 \text{ and } 10$
$\displaystyle 5 \text{ and } 6$
Use Tab to navigate between items. Click or tap a card to select it, then click or tap a card on the opposite side to create a match. You can also drag a card or its connector dot to a card on the opposite side.
Show the answer and how to solve it
The matches are $\sqrt{28} \to$ 5 and 6, $\sqrt{67} \to$ 8 and 9, and $\sqrt{150} \to$ 12 and 13. The pair 9 and 10 is unused.
1. 1For $\sqrt{28}$ , find the perfect squares on each side of 28. Since $5^{2} = 25$ and $6^{2} = 36$ , we have $5 < \sqrt{28} < 6$ , so it matches the pair 5 and 6.
2. 2For $\sqrt{67}$ , the surrounding perfect squares are $8^{2} = 64$ and $9^{2} = 81$ , so $8 < \sqrt{67} < 9$ , matching the pair 8 and 9.
3. 3For $\sqrt{150}$ , the surrounding perfect squares are $12^{2} = 144$ and $13^{2} = 169$ , so $12 < \sqrt{150} < 13$ , matching the pair 12 and 13. The pair 9 and 10 is unused.
Common mistake. A common mistake is matching $\sqrt{67}$ with 9 and 10. Since $9^{2} = 81 > 67$ , the value $\sqrt{67}$ is less than 9, so the correct pair is 8 and 9, not 9 and 10.
The idea underneath. If a number is between two perfect squares, then its square root is between the corresponding integers.
3On level
Look at Emma's comparison work. Which statement best describes the work?
Step 1
$\displaystyle 3 < \sqrt{14} < 4$
Step 2
$\displaystyle 3 < \sqrt{15} < 4$
Conclusion
$\displaystyle \sqrt{14} = \sqrt{15}$
A
The student forgot that negatives reverse the order.
B
No mistake; the comparison is correct.
C
The student used bounds that are too wide to prove the comparison.
D
The student rounded too much and needed more decimal places.
E
The student compared what was inside a square root instead of the irrational numbers.
Show the answer and how to solve it
The best description is that the student used bounds that are too wide to prove the comparison. Knowing both numbers are between 3 and 4 does not show they are equal.
1. 1Emma showed that both $\sqrt{14}$ and $\sqrt{15}$ lie somewhere between 3 and 4.
2. 2The two intervals are identical: [3, 4] for both numbers. Completely overlapping bounds only confirm that both values share a region—they say nothing about where in that region each one falls, so they cannot prove equality.
3. 3Use closer decimal approximations to separate the numbers.
4. 4Now the values separate. Since 3.74 < 3.87, we have $\sqrt{14} < \sqrt{15}$ . Emma's reasoning error was using bounds that were too wide to justify the conclusion $\sqrt{14} = \sqrt{15}$ .
Common mistake. A common mistake is assuming that two irrational numbers sharing the interval [3, 4] must be equal. That interval contains infinitely many different values, so tighter bounds or more precise decimal approximations are needed to determine the comparison.
The idea underneath. To compare irrationals, the bounds or approximations must separate the numbers; overlapping bounds only narrow the region, not the order.
4Stretch
Which interval on the number line contains this number?
$\displaystyle -\sqrt{5}$
A
-2.4 to -2.3
B
-2.3 to -2.2
C
-2.2 to -2.1
D
2.2 to 2.3
Show the answer and how to solve it
The answer is $-2.3$ to $-2.2$ because $2.2^{2} = 4.84$ and $2.3^{2} = 5.29$ , so $-\sqrt{5}$ lies between those tenths.
1. 1Look at the number shown: $-\sqrt{5}$ .
2. 2First bound $\sqrt{5}$ by checking squares of nearby tenths.
3. 3Since $4.84 < 5 < 5.29$ , we know $2.2 < \sqrt{5} < 2.3$ . Negating every part of the inequality reverses it: $-2.3 < -\sqrt{5} < -2.2$ .
4. 4So the correct interval is $-2.3$ to $-2.2$ .
Common mistake. A common mistake is choosing $2.2$ to $2.3$ by correctly estimating $\sqrt{5}$ but forgetting the negative sign. Because $-\sqrt{5}$ is negative, its location is on the negative side of the number line.
The idea underneath. To place $\sqrt{n}$ on a tenths number line, compare $n$ to the squares of the tenths endpoints; the correct interval is the one whose endpoint squares trap $n$ .

Step 1	$\displaystyle 3 < \sqrt{14} < 4$
Step 2	$\displaystyle 3 < \sqrt{15} < 4$
Conclusion	$\displaystyle \sqrt{14} = \sqrt{15}$