Question

21952 square root without calculator

Answer 1

Step-by-step explanation:

there is a way to find approximation of non-perfect square root:

$\sqrt{x} \approx \: \frac{1}{2}(y + \frac{x}{y} )$

where $x$ is number you want to find square root of and $y$ is the closest approximation you can guess,

so lets solve it,

using long division method lets find the approximation of $\sqrt{21952}$ in whole number,

that is $148$.

so now we now the approximation in whole number,

therefore,

$x = 21952$

because it is the number of which we want to find square root.

$y = 148$

because it is an approximation of square of which we want to find square root.

so, now,

$\frac{ 1 }{ 2 } (148+ \frac{ 21952 }{ 148 } )$

$=\frac{1}{2}\times \left(\frac{5476+5488}{37}\right) \\ = \frac{1}{2}\times \left(\frac{10964}{37}\right) \\ = \frac{1\times 10964}{2\times 37} \\ = \frac{10964}{74} \\ = \frac{5482}{37} = 148.1621621...$

that is pretty close to actual square root that is,

$\sqrt{21952} = \: 148.1620734...$

And our approximation,

$\sqrt{21952} \approx \: 148.1621621...$

And to be more accurate repeat the process again,

now, as we got more accurate approximation of $\sqrt{21952}$ that is $\frac{5482}{37}$.

therefore, now $y = \frac{5482}{37}$

that is,

$\frac{ 1 }{ 2 } ( \frac{ 5482 }{ 37 } + \frac{ 21952 }{ \frac{ 5482 }{ 37 } } ) \\ = \frac{1}{2}\left(\frac{5482}{37}+21952\times \left(\frac{37}{5482}\right)\right) \\= \frac{1}{2}\left(\frac{5482}{37}+\frac{21952\times 37}{5482}\right) \\ = \frac{1}{2}\left(\frac{5482}{37}+\frac{812224}{5482}\right) \\ = \frac{1}{2}\left(\frac{5482}{37}+\frac{406112}{2741}\right) \\ = \frac{1}{2}\times \left(\frac{15026162+15026144}{101417}\right) \\ = \frac{1}{2}\times \left(\frac{30052306}{101417}\right) \\ = \frac{1\times 30052306}{2\times 101417} \\ = \frac{30052306}{202834} \\ = \frac{15026153}{101417} \: = 148.16207342...$

now we got more accurate value for $\sqrt{21952}$.

So, Hope it all helped.

Because it all is possible to be done by hand.

At last i want to say that the more accurate the $y$ is the more accurate your square root is.