Question

How do you evaluate `\sqrt{75^{2}-56^{2}}`?

Answer 1

`sqrt(75^2-56^2) = sqrt(2489) ~~ 49.89`

Explanation:

The difference of squares identity tells us:

`a^2-b^2 = (a-b)(a+b)`

So:

`sqrt(75^2-56^2) = sqrt((75-56)(75+56))`

`color(white)(sqrt(75^2-56^2)) = sqrt(19*131)`

`color(white)(sqrt(75^2-56^2)) = sqrt(2489)`

Both `19` and `131` are prime numbers, so this square root is in simplest form.

Since `2489` is close to `2500 = 50^2`, we can fairly quickly get good rational approximations to the irrational number `sqrt(2489)` using the Babylonian method, or by using generalised continued fractions.

For example:

`sqrt(a^2+b) = a+b/(2a+b/(2a+b/(2a+b/(2a+...))))`

So putting `a=50` and `b=2489-a^2 = -11` we get:

`sqrt(2489) = 50-11/(100-11/(100-11/(100-11/(100-...))))`

So:

`sqrt(2489) ~~ 50-11/100 = 50-0.11 = 49.89`