How do I approximate an irrational square root without using a calculator?

Although this might look like a smart trick at the beginning, the truth is that as a physicist this kind of approximations turn out to be useful very often. This is quite likely to come up in the PAT as well!

It is important to get used to understanding symbolic expressions (no numbers!), so I will first derive the maths and then apply the resulting expression to a couple of illustrative examples.

Our aim is to obtain an approximate value for the inexact square root x of an arbitrary number n, i.e. to solve x = n1/2. In other words, we want to solve the equation f(x) = x2 = n. Suppose we know the exact root x0 of another number n0 which is very close to n, i.e. f(x0) = x02 = n0 (this is key!). We can do a Taylor series expansion of f(x) about x0 up to first order (first derivative f'(x)), which is simply:

f(x) = f(x0) + f'(x0)(x - x0).

The first derivative f'(x) = 2x, which evaluated at x0 is f'(x0) = 2x0. Substituting f(x) = n and f(x0) = n0 leads to:

n = n0 + 2x0(x - x0).

We are almost done! We know all the values of n, n0 and x0 and just need to solve for x. Rearranging:

x = x0 + (n - n0)/(2x0)

That's it! Remember that this expression is just an approximation for the actual square root of n, but a quite good one the closer n0 is to n.

Let's plug in some numbers. Suppose you were asked to approximate the square root of n = 66. Despite the solution being irrational, there is a close number n0 = 64 with a nice exact square root x0 = 8. Using the expression above we find:

x = 8 + (66 - 64)/(2*8) = 8 + 2/16 = 8 + 1/8 = 8 + 0.125 = 8.125

How good is this? It's just 0.01% off the actual value 8.12404...!

What if we chose n0 = 81 instead, the square root of which is x0 = 9? This would give x = 8.167 - worse, but still a reasonable result considering how far 81 is from 66. Never forget to quote both the positive and negative square roots in your final solution!

Would you be able to derive a similar expression for approximating a cubic root?

Sergio H. Uni Admissions Test .PAT. tutor, IB Spanish tutor, IB Maths...

2 years ago

Answered by Sergio, an Uni Admissions Test .PAT. tutor with MyTutor

Still stuck? Get one-to-one help from a personally interviewed subject specialist


£26 /hr

Kirill M.

Degree: Physics (Masters) - Oxford, St Catherine's College University

Subjects offered:.PAT., Physics+ 2 more

Further Mathematics

“Me Hi, I'm Kirill, I'm a 3rd year physicist at Oxford. Whether you are looking for exam preparation or for a better understanding of the material that you are covering in your studies of Maths or Physics, I can help you reach your goa...”

£30 /hr

Alex S.

Degree: Physics (Bachelors) - Oxford, St Peter's College University

Subjects offered:.PAT., Physics+ 2 more

Further Mathematics

“Studying Physics at the University of Oxford. Looking to tutor maths and physics at all levels, be sure to send me a message if you have any questions!”

£25 /hr

Edward C.

Degree: Physics (Masters) - Oxford, St Anne's College University

Subjects offered:.PAT., Physics+ 1 more


“I am currently studying Physics at the University of Oxford. I am really passionate about my subject and enjoy teaching it as a result.”

MyTutor guarantee

About the author

Sergio H.

Currently unavailable: until 01/08/2015

Degree: Physics with Theoretical Physics (Bachelors) - Imperial College London University

Subjects offered:.PAT., Spanish+ 2 more


“Third year Physics student at Imperial College. Looking to help you with the two fields I love, Physics and Maths, and with Spanish as a native speaker.”

MyTutor guarantee

You may also like...

Other Uni Admissions Test .PAT. questions

How do I approximate an irrational square root without using a calculator?

How do I evaluate something like 3070^2-3069^2?

How are you to find 2007^2 − 2006^2 without a calculator?

How many trailing zeros are there in 100! (100 factorial)?

View Uni Admissions Test .PAT. tutors

We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this. Dismiss