What is the square root of 26?

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What is the square root of 26?

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Answer 1 · 2015-09-06T21:00:57

You can only have an approximation: 5.09901951...

Explanation:

The square root of a number $x$ $x$ is a number $y$ $y$ such that $y^{2} = x$ $y^2=x$ . So, we're looking for a number $y$ $y$ such that $y^{2} = 26$ $y^2=26$ . Since $5^{2} = 25$ $5^2=25$ and $6^{2} = 36$ $6^2=36$ , the square root of $26$ $26$ is between $5$ $5$ and $6$ $6$ . Since there is no algorithm to compute it exactly, you can only have an approximation. A possible way is the following: we know that $\sqrt{26}$ $sqrt(26)$ is between $5$ $5$ and $6$ $6$ . So, since $5^{2} = 25$ $5^2=25$ and ${5.1}^{2} = 26.01$ $5.1^2=26.01$ , $\sqrt{26}$ $sqrt(26)$ must be between $5$ $5$ and $5.1$ $5.1$ .

Iterating this process gives you all the decimal digits you need.

Answer 2 · 2015-09-06T23:14:22

$\sqrt{26}$ $sqrt(26)$ does not simplify, but you can calculate an approximation efficiently using Newton Raphson method as:

$\sqrt{26} \approx \frac{54100801}{10610040} \approx 5.099019513592786$ $sqrt(26) ~~ 54100801 / 10610040 ~~ 5.099019513592786$

Explanation:

$26 = 2 \cdot 13$ $26 = 2 * 13$ has no square factors, so $\sqrt{26}$ $sqrt(26)$ cannot be simplified.

If you want to calculate an approximation by hand, then I would recommend a form of Newton Raphson method, starting with first approximation $a_{0} = 5$ $a_0 = 5$ .

To iterate you can use the formula:

$a_{i + 1} = \frac{a_{i}^{2} + n}{2 a_{i}}$ $a_(i+1) = (a_i^2 + n)/(2a_i)$

where $n = 26$ $n = 26$ is the number you are approximating the square root of.

Personally, I like to deal with these approximations as rational approximations in the form $\frac{p_{i}}{q_{i}} = a_{i}$ $p_i/q_i = a_i$ where $p_{i}$ $p_i$ and $q_{i}$ $q_i$ are integers as follows:

$n = 26$ $n = 26$
$p_{0} = 5$ $p_0 = 5$
$q_{0} = 1$ $q_0 = 1$

Iterate using:

$p_{i + 1} = p_{i}^{2} + n q_{i}^{2}$ $p_(i+1) = p_i^2 + n q_i^2$
$q_{i + 1} = 2 p_{i} q_{i}$ $q_(i+1) = 2 p_i q_i$

So:

$p_{1} = 5^{2} + 26 \cdot 1^{2} = 25 + 26 = 51$ $p_1 = 5^2 + 26*1^2 = 25+26 = 51$
$q_{1} = 2 \cdot 5 \cdot 1 = 10$ $q_1 = 2*5*1 = 10$

$p_{2} = 51^{2} + 26 \cdot 10^{2} = 2601 + 2600 = 5201$ $p_2 = 51^2 + 26*10^2 = 2601 + 2600 = 5201$
$q_{2} = 2 \cdot 51 \cdot 10 = 1020$ $q_2 = 2*51*10 = 1020$

$p_{3} = 5201^{2} + 26 \cdot 1020^{2} = 27050401 + 27050400 = 54100801$ $p_3 = 5201^2 + 26*1020^2 = 27050401 + 27050400 = 54100801$
$q_{3} = 2 \cdot 5201 \cdot 1020 = 10610040$ $q_3 = 2*5201*1020 = 10610040$

Stop when you think you have enough significant digits (typically about the number of significant digits of $p_{i}$ $p_i$ + the number of significant digits of $q_{i}$ $q_i$ ).

$\sqrt{26} \approx \frac{54100801}{10610040} \approx 5.099019513592786$ $sqrt(26) ~~ 54100801 / 10610040 ~~ 5.099019513592786$

Actually $\sqrt{26} \approx 5.099019513592785$ $sqrt(26) ~~ 5.099019513592785$

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What is the square root of 26?

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