Irrational numbers like $\sqrt{50}$ have decimals that never end, but you can still find where they sit, closely and without a calculator.
Why it works
For positive numbers, squaring keeps the order: if $a < b$, then $a^{2} < b^{2}$. So to place $\sqrt{50}$, find the perfect squares on each side of 50: $49 < 50 < 64$. Their roots are 7 and 8, so $7 < \sqrt{50} < 8$. And because 50 is just above 49, $\sqrt{50}$ is just above 7.
Key idea
To estimate $\sqrt{n}$, find the perfect squares just below and just above $n$; the root lies between their roots. To get closer, square some tenths and see which one lands nearest $n$.
Worked examples
Worked example: Between two whole numbers
Estimate $\sqrt{50}$ to the nearest tenth.
- $7^{2} = 49$ and $8^{2} = 64$, so $\sqrt{50}$ is a bit more than 7.
- Try tenths: $7.0^{2} = 49$ and $7.1^{2} = 50.41$.
- 50 is between them and closer to 50.41 (0.41 away, versus 1 away from 49).
Answer: $\sqrt{50} \approx 7.1$
Worked example: Compare with π
Which is greater, $\sqrt{10}$ or $\pi$?
- $3.1^{2} = 9.61$ and $3.2^{2} = 10.24$, so $\sqrt{10}$ is between 3.1 and 3.2.
- $3.15^{2} = 9.9225$ is still less than 10, so $\sqrt{10} > 3.15$.
- $\pi \approx 3.14$, so $\sqrt{10}$ is a little greater.
Answer: $\sqrt{10}$
Worked example: Put them in order
Order from least to greatest: $\sqrt{30}$, $5.6$, $\frac{17}{3}$.
- $5.4^{2} = 29.16$ and $5.5^{2} = 30.25$, so $\sqrt{30}$ is between 5.4 and 5.5.
- $\frac{17}{3} = 5.\overline{6}$, a little more than 5.6.
Answer: $\sqrt{30} < 5.6 < \frac{17}{3}$
Common mistake
Halving: estimating $\sqrt{50}$ as 25. Check by squaring: $25^{2} = 625$, far more than 50. Look for the number whose square is near 50 instead.