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# Estimating Square Roots

## Task

Without using the square root button on your calculator, estimate $\sqrt{800}$ as accurately as possible to $2$ decimal places.

(Hint: It is worth noting that $20^2 = 400$ and $30^2=900$.)

## IM Commentary

By definition, the square root of a number $n$ is the number you square to get $n$. The purpose of this task is to have students use the meaning of a square root to find a decimal approximation of a square root of a non-square integer. Students may need guidance in thinking about how to approach the task.

## Solutions

Solution: Using the definition of a square root

We know that $$20^2=400$$ and $$ 30^2=900$$ so $$20 \lt \sqrt{800} \lt 30$$

Choosing successive approximations carefully, we see that:

$n$ | $n^2$ | $m^2$ | $m$ |
---|---|---|---|

28 | 784 | 851 | 29 |

28.2 | 795.24 | 800.89 | 28.3 |

28.28 | 799.7584 | 800.3241 | 28.29 |

28.284 | 799.984656 | 800.041225 | 28.285 |

So $\sqrt{800} \approx 28.28$.

Solution: Another approach

We know that $20^2=400$ and $ 30^2=900,$ so $$20 \lt \sqrt{800} \lt 30.$$

If we take the average of 20 and 30, we get $\frac{20+30}{2} = 25$. Since $25^2 = 625$, we know that

$$25 \lt \sqrt{800} \lt 30.$$

If we take the average of 25 and 30, we get $\frac{25+30}{2} = 27.5$. Since $27.5^2 = 756.25$, we know that

$$27.5 \lt \sqrt{800} \lt 30.$$

If we take the average of 27.5 and 30, we get $\frac{27.5+30}{2} = 28.75$. Since $28.75^2 = 826.5625$, we know that

$$27.5 \lt \sqrt{800} \lt 28.75.$$ Continuing in this way, we get $$\sqrt{800} \approx 28.28.$$

## Estimating Square Roots

Without using the square root button on your calculator, estimate $\sqrt{800}$ as accurately as possible to $2$ decimal places.

(Hint: It is worth noting that $20^2 = 400$ and $30^2=900$.)

## Comments

Log in to comment## yuchun says:

over 3 yearswe can also use more quickly method to get the results, if you don't mind, I can post my solution here..

## Cam says:

over 3 yearsOf course, you're more than welcome to post your solution here.

## hollandr says:

over 5 yearsJust an editing comment. In the second level of the approximations table, the last column, m, should read 28.3, not 29.3.

## Cam says:

over 5 yearsFixed. Thanks!