$\begingroup$

The square root of $64$ is $8$.

If the decimal of $64$ is shifted by two places to the left (i.e. $0.64$), then the decimal of the answer is shifted by one place to the left (i.e. $0.8$), so the square root of $0.64$ is $0.8$.

Is there any intuitive way to think about the case where the decimal is shifted by only one place? i.e. square root of $6.4$, which is $2.53$.

$\endgroup$ 1

2 Answers

$\begingroup$

This rule works only when the radicand may be expressed as $a^2\times 10^{2n},$ where $a,n$ are integers, and from this expression it is obvious to see why it works.

For the other case (of which it is sufficient to consider your example without loss of generality), the rule here is radically different (no pun intended). For example, we have that $$\sqrt{6.4}=\sqrt{\frac{64}{10}}=\frac{8}{\sqrt {10}}=0.8\sqrt{10}.$$ In all such cases we can always reduce the radical to the form $b\times \sqrt {10},$ where $b$ is a rational number. So you only need to memorise a convenient rational approximation to $\sqrt {10}$ if you want to be able to approximate such radicals quickly to some desired degree of accuracy.

$\endgroup$ $\begingroup$

Simply $\sqrt {10x} = \sqrt {10}\cdot \sqrt x$

And $\sqrt {10}\approx 3.16$

That should also help you to think through any similar problems/thoughts you may have.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy