Explain why the square root of 89 must be between 9 and 10?

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Explain why the square root of 89 must be between 9 and 10?

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Answer 1 · 2016-06-01T18:54:48

Evaluating the square of $9$ and $10$ .

Explanation:

Recall that the square root is the inverse operation of the square.
Then we can see that the square of $9$ is

$9^2=81$

and the square of $10$ is

$10^2=100$

comparing with $89$ we have that $89>81$ and $89<100$ .
Doing the square root we have

$sqrt(89) > sqrt(81)$ that is $sqrt(89)>9$

$sqrt(89) < sqrt(100)$ that is $sqrt(89)<10$ .
We proved that the square root of $89$ is greater than $9$ and smaller than $10$ .

Answer 2 · 2016-06-01T18:55:18

$9 < sqrt(89) < 10$

Explanation:

We have $81 = 9^2 < 89 < 10^2 = 100$
and $sqrt(x)$ is monotonic for $0 < x < infty$
which means that if $0 < x_1 < x_2->sqrt(x_1) < sqrt(x_2)$
then $9 < sqrt(89) < 10$

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Explain why the square root of 89 must be between 9 and 10?

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