A state department of education, as a question on the public-release form of a statewide standardized test, released a question to the public in which eighth-grade students have to locate the square root of 10 on a number line.
The square root of 10 is an irrational number, meaning it cannot be represented as a fraction: its decimals do not repeat.
On a calculator, it comes out to 3.1622776601683793319988935444327…..
That is just a little greater than 3 but less than 4, and since only one of the choices on the number line is higher than 3 (to the right of 3) and lower than 4 (to the left of 4), that one has to be the answer.
The other dots are at about 0.9, which is less than 1 and therefore can’t be the answer; 2.6, which is less than 3, not greater than 3; and at 5, which is the square root of 25, not of 10.
The steps: First, memorize all the perfect squares from 1 to 15, and then memorize those for the 5’s after that. Here they are:
| Number | Perfect Square |
| 1 | 1 |
| 2 | 4 |
| 3 | 9 |
| 4 | 16 |
| 5 | 25 |
| 6 | 36 |
| 7 | 49 |
| 8 | 64 |
| 9 | 81 |
| 10 | 100 |
| 11 | 121 |
| 12 | 144 |
| 13 | 169 |
| 14 | 196 |
| 15 | 225 |
| 20 | 400 |
| 25 | 625 |
| 30 | 900 |
| 35 | 1,225 |
| 40 | 1,600 |
Once you have those memorized, then you will be able to estimate the square root of any number that is put on a standardized test.
For example, if you are asked to estimate the square root of 200, you know the square root of 225 is 15, and the square root of 196 is 14. Therefore, the square root of 200 will be in between 14 and 15, and probably closer to 14 than to 15, since 200 is closer to 196 than it is to 225.
Locating a decimal number on a number line is a lower-level skill than estimating the square root of a number, and we will discuss this skill in sixth-grade questions.
Want to get it exactly right?
You can find the exact place where the square root of 2 is located on the number line using a compass and the Pythagorean Theorem.
First, notice that the square root of 2 is the length of the hypotenuse of an isosceles right triangle with legs of length 1. Using the number line, set the compass to a width of 1 (put the point at 0 and the pencil on 1, and lock it).
Next, construct a line segment of length 1 perpendicular to the number line, starting at 1 and going straight up. We have left out the construction of the line segment, since it makes the construction image somewhat busy. We’d rather focus on the isosceles right triangle, the key to our diagram.
Next, connect the top of your perpendicular line segment to the point 0 on the number line. This hypotenuse has a length equal to the square root of 2 on this number line.
Finally, set the width of the compass to the length of the hypotenuse, put the point on 0, and construct an arc that intersects the number line. The point where your arc intersects the number line will be the exact square root of 2.
You can find other square roots using a similar technique.
For example, note that the square root of 10 is the length of the hypotenuse of a right triangle with legs of length 1 and 3. And the square root of 11 is the hypotenuse of a right triangle with legs measuring 1 and the square root of 10. That is, once you get the square root of 10 in the first part of this example, you can easily get the square root of 11 and many other square roots located on the number line.
The Answers
1. The square root of 90 is about 9.4868. Without a calculator, though, you should know that it’s about half-way between 9 and 10. I say that because 9 is the square root of 81, which is 9 less than 90, and 10 is the square root of 100, which is 10 more than 90. The problem asked for an estimate of the square root of 90, which is a teeny, tiny bit closer to the square of 9 (81) than it is to the square of 10 (100), so the square root of 90 is in between 9 and 10, but just a little bit closer to 9 than to 10.
Regarding how to locate this estimate on a number line, first we draw a number line that includes both 9 and 10. On that number line, make a little mark (shown in green) about halfway in between the 9 and the 10:
Since our point is closer to 9 than to 10—but not by much—place it just below (to the left of) the halfway mark (shown in red):
2. The square root of 125 is 11.1803. It’s very close to 11, since 11 is the square root of 121, which is only 4 numbers away from 125, which is what the problem asked for. You have to go all the way up to 144 to get the square of 12, and that number is much farther away from 125 than 121 is. Therefore, we would estimate that the square root of 125 is in between 11 and 12, but that it is closer to 11 than it is to 12.
To locate our estimate on the number line, we notice that 125 is 4 greater than 121, and that the total distance from 121 (the square of 11) to the next perfect square (144, the square of 12) is 23 units. Therefore our point should be estimated at about 4/23 of the way from 11 to 12. To make things easier, we can fudge both the numerator and denominator of this fraction upwards—to 5/25, which is equivalent to 1/5. That means our estimate is pretty good if we plot the point that represents the square root of 125 one-fifth of the way between 11 and 12. Divide the space between 11 and 12 into approximate fifths:
Now, since we fudged up the denominator by two to 25 (a fudge factor of 2/23) and the numerator by one to 5 (representing a larger fudge factor of 1/4), that means our estimate should be just a little bit below where we had planned. That is, 4/23 is less than 5/25, because the increase in the numerator was greater than the increase in the denominator, resulting in a fraction that is greater than the original 4/23. Locate the point just to the left of the 1/5 tick mark:
It would also have been acceptable to estimate 4/23 to 4/24, which would reduce to 1/6, but it is easier for me to divide a space on the number line into fifths than it is for me to divide it into sixths for some reason (just a little visually challenged, I suppose). If you were to estimate 4/23 to 4/24, bumping only the denominator up, you would end up with a fraction that’s smaller than your original 4/23. Therefore, you would plot the estimated square root of 125 to the right (just by a very small amount) of the 1/6 mark.
3. The square root of 300 is about 17.3205. It is an irrational number, like all the other ones we’ve been doing here. We know that the square of 15 is 225 and the square of 20 is 400. Our problem, 300, is just a little closer to 225 (75 away) than it is to 400 (100 away), but we also know that squares start increasing at a greater rate as the square roots get higher. Therefore, even though there is a 25-point difference between the two ends, the actual square root ends up being rather close to the midpoint between 15 and 20. So, the lesson to learn here is that we have to “fudge” our answer a little when we only know the perfect squares of numbers that are 5 or more away from the one asked for in the problem. So, we might have added (3/7) x (5) to 15 in order to obtain an initial estimate, since 300 is 3/7ths of the way from 225 to 400. That would have given us 2.14… and an estimate of 17.14. But since we are fudging higher, 17.32 is probably not a bad estimate.
4. The square root of 2 is 1.4142 (it rhymes, so you could make up a little song … “The square root of two … is 1 point 4-1-4-2 … How do you do? … And I like you” …). In fact, you may want to memorize that little math fact as well. It is a very common number, and if you work better in decimals than with radicals hanging around, you might as well just commit the square root of 2’s estimated decimal value to memory. Clearly the 2 is in between 1 and 4, and therefore the square root of 2 would have to be in between 1 and 2. And so it is.
Try a few on your own.
1. Estimate the square root of 90.
2. Estimate the square root of 125.
3. Estimate the square root of 300.
4. Estimate the square root of 2.
5. Just pick a random number between 1 and 999, and estimate its square root. Then plot that square root on a number line. Practice as many as you need to in order to master this skill.