How do you find the square root of 52?

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How do you find the square root of 52?

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Answer 1 · 2015-09-12T20:52:18

$\sqrt{52} = 2 \sqrt{13} \approx 7.21110$ $sqrt(52) = 2sqrt(13) ~~ 7.21110$

Explanation:

If $a, b \geq 0$ $a, b >= 0$ then $\sqrt{a b} = \sqrt{a} \sqrt{b}$ $sqrt(ab) = sqrt(a)sqrt(b)$ , so:

$\sqrt{52} = \sqrt{2^{2} \cdot 13} = \sqrt{2^{2}} \sqrt{13} = 2 \sqrt{13}$ $sqrt(52) = sqrt(2^2*13) = sqrt(2^2)sqrt(13) = 2sqrt(13)$

If you want to calculate an approximation by hand, you could follow the advice I gave for $\sqrt{28}$ $sqrt(28)$ in http://socratic.org/questions/how-do-you-find-the-square-root-28

Using the method described there:

Let $n = 52$ $n = 52$ , $p_{0} = 7$ $p_0 = 7$ , $q_{0} = 1$ $q_0 = 1$

Then:

$p_{1} = 7^{2} + 52 \cdot 1^{2} = 49 + 52 = 101$ $p_1 = 7^2+52*1^2 = 49+52 = 101$
$q_{1} = 2 \cdot 7 \cdot 1 = 14$ $q_1 = 2*7*1 = 14$

$p_{2} = 101^{2} + 52 \cdot 14^{2} = 10201 + 10192 = 20393$ $p_2 = 101^2+52*14^2 = 10201+10192 = 20393$
$q_{2} = 2 \cdot 101 \cdot 14 = 2828$ $q_2 = 2*101*14 = 2828$

Stopping at this point, we get a result accurate to $5$ $5$ decimal places:

$\sqrt{52} \approx \frac{20393}{2828} \approx 7.21110$ $sqrt(52) ~~ 20393/2828 ~~ 7.21110$

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How do you find the square root of 52?

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