Today I want to do some math exercises and suddenly I found that it asks me how much is radical of 124. However, I have made some researches and it gave me a very long number which is 11, and with huge numbers which is totally wrong.
How can I calculate in easy way? Or, is there a easy way to do this?
We have $124=2^2\cdot 31$, hence the radical of $124$ is $$ {\rm rad}(124)=2\cdot 31=62. $$
Reference: Wikipedia.
Edit: Due to translation error, it seems that you mean $$ \sqrt{124}=11.135528725660043844238942597837099041\cdots $$
What counts as an easy way depends on how hard you want to work and how accurate an answer you need.
Clearly, since $11^2 = 121$ the answer will be a little bit larger than $11$.
You could experiment with $11.1$, $11.05$ and so on to get closer.
There's an approximation you learn in calculus that says $$ \sqrt{x+h} \approx \sqrt{x} + \frac{h}{2\sqrt{x}}. $$ when $h$ is small compared to $x$, so $$ \sqrt{121 + 3} \approx 11 + 3/22. $$
You can check: $$ (11 + 3/22)^2 = 124.018595041... $$
However, I have made some researches and it gave me a very long number which is 11, and with huge numbers which is totally wrong.
That's totally right. $11^2 =121 < 124 < 144 = 12^2$ so $11 < \sqrt{124} < 12$. $(11.25)=126.5625 > 124$ so $11 < \sqrt{124} < 11.25$. $11.1^2 = 123.21$ so $11.1 < \sqrt{124} < 11.25$ and so on.
Now $124$ is not a perfect square and so the square root will not be rational so we can estimate this forever and ever and never end and we'll only get an estimation. And we live in the 21st century so if there's nothing to be learned by doing it by hand we can use a calculator and get $\sqrt{124} \approx 11.135528725660043844238942597837$.
But that assumes we care want value it is close to. If you are doing a practical problem that requires us to buy enough paint to paint a square with the are of $124$ sq ft or whatever we can estimate to as much accuracy as we want.
But if we are theoretical mathematicians (yeah, go team!) we actually don't care in the least as to what size the square root of one-hundred-twenty-four is. We care about writing an expression, $blah$, so that it $blah$ is an express that we know $blah^2 = 124$ (and $blah$ is positive). AND we want the expression to be must versatile so that $blah$ has other properties (such as it is twice the square root of something else or whatever we can see that as well.)
$\sqrt{124}$ is such an expression but we'd like to simplify it so it doesn't have any square factors only the radical sign. i.e We'd like to express it as $m\sqrt{n}$ so that $n$ doesn't have any square in it.
So to do that we figure out the prime factorization of $124$.
$124 = 2*62 = 2*2*31$ and $31$ is prime so $124 = 2^2*31$.
So $\sqrt{124} = \sqrt{2^2*31} = \sqrt{2^2}*\sqrt{31} = 2\sqrt{31}$. (And we know $\sqrt{31}$ can't be simplified further because $31$ is not a perfect square and so $\sqrt{31}$ is not rational and can not be expressed as a fraction between integers or as a decimal that terminates or has a periodal repitition.)
Another example:
What is $\sqrt{85584600}$?
$85584600= 2*42792300 = 2*2*21396150=2*2*2*10698075=$
$2^3*3*10698075=2^3*3*3*1188675=2^3*3^3*396225=2^3*3^4*132075=2^3*3^5*44025=2^3*3^6*14675=$
$2^3*3^6*5*2935= 2^3*3^6*5^2*587$ and $587$ is prime
So $\sqrt{42792300}= \sqrt{2^3*3^6*5^2*587}=\sqrt{(2^2*3^6*5^2)*(2*587)}=$
$\sqrt{(2*3^3*5)^2*(2*587)}=\sqrt{(2*3^3*5)^2}*\sqrt{2*587}=$
$2*3^3*5\sqrt{1174}= 270\sqrt{1174}$.
This is Ethan Bolker solution in more details:
$f'(x)=\frac{f(x+\Delta x)-f(x)}{\Delta x}$
$f(x+\Delta x)≈f(x)+f'(x).\Delta x$
We use function $f(x)=\sqrt x$ ; we have:
$x+\Delta x=124=121+3$ ⇒ $x=121$ and $\Delta x =3$
$f'(x)=\frac{1}{2\sqrt x}$
puting values of x and $\Delta x$ we get:
$\sqrt {124}=\sqrt{121}+\frac{1}{2\sqrt {121}}.3=11+\frac{3}{22}≈11.13636...$
