How do you find the square root of 20?

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How do you find the square root of 20?

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Answer 1 · 2016-08-31T20:59:55

Find approximation:

$\sqrt{20} \approx \frac{2889}{646} \approx 4.472136$ $sqrt(20) ~~ 2889/646 ~~ 4.472136$

Explanation:

The prime factorisation is:

$20 = 2^{2} \cdot 5$ $20 = 2^2*5$

Hence:

$\sqrt{20} = 2 \sqrt{5}$ $sqrt(20) = 2sqrt(5)$

This is an irrational number between $4$ $4$ and $5$ $5$ since:

$4^{2} = 16 < 20 < 25 = 5^{2}$ $4^2 = 16 < 20 < 25 = 5^2$

It is not expressible as an exact fraction, but we can find rational approximations...

Since $20$ $20$ is roughly halfway between $4^{2}$ $4^2$ and $5^{2}$ $5^2$ , its square root is approximately $\frac{9}{2}$ $9/2$ - halfway between $4$ $4$ and $5$ $5$ .

In fact, we find:

$9^{2} = 81 = 80 + 1 = 20 \cdot 2^{2} + 1$ $9^2 = 81 = 80+1 = 20*2^2 + 1$

which is in Pell's equation form:

$p^{2} = n q^{2} + 1$ $p^2 = n q^2 + 1$

with $n = 20$ $n = 20$ , $p = 9$ $p = 9$ and $q = 2$ $q = 2$

That means that we can deduce the continued fraction for $\sqrt{20}$ $sqrt(20)$ from the continued fraction for $\frac{9}{2}$ $9/2$ ...

$\frac{9}{2} = 4 + \frac{1}{2} = [4; 2]$ $9/2 = 4+1/2 = [4;2]$

Hence:

$sqrt(20) = [4;bar(2,8)] = 4+1/(2+1/(8+1/(2+1/(8+1/(2+1/(8+...))))))$

To get a good approximation for $sqrt(20)$ truncate this continued fraction early, just before an ' $8$ '...

For example:

$sqrt(20) ~~ [4;2,8,2,8,2] = 4+1/(2+1/(8+1/(2+1/(8+1/2)))) = 2889/646 ~~ 4.472136$

From a calculator:

$sqrt(20) ~~ 4.472135955$

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How do you find the square root of 20?

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