How do you find the square root of 16562?

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How do you find the square root of 16562?

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Answer 1 · 2016-06-08T12:50:55

$\sqrt{16562} \approx 128.699$ $sqrt(16562) approx 128.699$

Explanation:

The ancient greeks used to compute square roots by sucessive approximations.

Given a number $N$ $N$ for which they need to compute the square root and given an initial approximation $q_{0}$ $q_0$ they proceed as follows:

${(q_{0} + δ q_{0})}_{0}^{2} = N$ $(q_0+ delta q_0)^2=N$ or
$q_{0} + 2 q_{0} δ q + {(δ q_{0})}_{0}^{2} = N$ $q_0+2 q_0 delta q + (delta q_0)^2 = N$

So they were searching for an approximation variation $δ q_{0}$ $delta q_0$ with the purpose of correct the initial guess $q_{0}$ $q_0$ . They supposed also that $δ q_{0}$ $delta q_0$ being small, much smaller would be ${(δ q_{0})}_{0}^{2}$ $(delta q_0)^2$ so they used the approximation

$q_{0} + 2 q_{0} δ q_{0} \approx N$ $q_0+2 q_0 delta q_0 approx N$

solving for $δ q_{0}$ $delta q_0$ they got

$δ q_{0} = \frac{(\frac{N}{q_{0}}) - q_{0}}{2}$ $delta q_0 = ((N/q_0) - q_0)/2$

once corrected $q_{0}$ $q_0$ they got $q_{1} = q_{0} + δ q_{0}$ $q_1 = q_0 + delta q_0$
and then follow with

$δ q_{1} = \frac{(\frac{N}{q_{1}}) - q_{1}}{2}$ $delta q_1 = ((N/q_1) - q_1)/2$ etc.

Let us apply that process for calculation of square root of

$N = 16562$ $N =16562$

our initial guess will be $q_{0} = 400$ $q_0 = 400$

$δ q_{0} = \frac{(\frac{16562}{400}) - 400}{2} = - 179.298$ $delta q_0 =((16562/400)-400)/2 =-179.298$

so $q_{1} = 400 - 179.298$ $q_1 = 400-179.298$ and calculating $δ q_{1}$ $delta q_1$

$δ q_{1} = - 72.83015$ $delta q_1=-72.83015$ so $q_{2} = 400 - 179.298 - 72.83015$ $q_2 = 400-179.298-72.83015$

calculating $δ q_{2}$ $delta q_2$

$δ q_{2} = - 17.93516$ $delta q_2 = -17.93516$ so $q_{3} = 400 - 179.298 - 72.83015 - 17.93516$ $q_3 = 400-179.298-72.83015-17.93516$

In the third iteration we get

$δ_{3} = - 1.23779$ $delta_3 = -1.23779$ obtaining a result of

$\sqrt{16562} \approx 128.699$ $sqrt(16562) approx 128.699$ which is a satisfactory result.

Answer 2 · 2016-06-08T12:51:25

$\sqrt{16562} = 91 \sqrt{2} = 128.6922$ $sqrt16562=91sqrt2=128.6922$

Explanation:

To find square root of $16562$ $16562$ , we should first factorize it.

From divisibility rules, it is apparent that it is divisible by $2$ $2$ and dividing by $2$ $2$ , we get $8281$ $8281$ .

$8281$ $8281$ is clearly not divisible by $3$ $3$ and $5$ $5$ , but is divisible by $7$ $7$ . Dividing by $7$ $7$ , we get $1183$ $1183$ , which is again divisible by $7$ $7$ and dividing it by $7$ $7$ we get $169$ $169$ , which is $13 \times 13$ $13xx13$ .

Hence, $16562 = 2 \times 7 \times 7 \times 13 \times 13$ $16562=2xx7xx7xx13xx13$ and hence

$\sqrt{16562} = \sqrt{2 \times ¯ ¯¯¯¯¯¯¯ ¯ 7 \times 7 \times ¯ ¯¯¯¯¯¯¯¯¯¯¯ ¯ 13 \times 13}$ $sqrt16562=sqrt(2xxbar(7xx7)xxbar(13xx13))$

= $7 \times 13 \times \sqrt{2} = 91 \sqrt{2} = 91 \times 1.4142 = 128.6922$ $7xx13xxsqrt2=91sqrt2=91xx1.4142=128.6922$

Answer 3 · 2016-06-09T09:57:48

$128.69$ $128.69$ to 2 decimal places

Explanation:

If you are not sure of the numbers build a factor tree to find the squared prime numbers.

Tont B

Thus $\sqrt{16562} = \sqrt{2 \times 7^{2} \times 13^{2}}$ $sqrt(16562) =sqrt(2xx7^2xx13^2)$

$= 7 \times 13 \times \sqrt{2} = 91 \sqrt{2} \approx 128.69$ $=7xx13xxsqrt(2) = 91sqrt(2)~~128.69$ to 2 decimal places

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How do you find the square root of 16562?

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