Question

How do you find a one-decimal place approximation for `root3 99`?

Answer 1

`4.6` and `4.7`.

Explanation:

If `root(3)(99)` is between `a` and `b`, then the following must hold:

`a^3 < 99 < b^3`.

So, we must look for two numbers `a` and `b` with this property, and they must be less than `1/10` apart.

First of all, let's focus on the nearest integer: we only need to make some calculation and go on with trials and errors: we'll begin listing the first cubes:

• `1^3 = 1`;
• `2^3 = 8`;
• `3^3 = 27`;
• `4^3 = 64`
• `5^3 = 125`.

From this list, we deduce that the third root of `99` is between `4` and `5`.

Now we can simply list the numbers between `4` and `5` with one decimal digit, and compute their cubes (which may be boring, but we can easily do it without a calculator, so it's not "cheating"):

• `4,1^3=68.921`;
• `4,2^3=74.088`;
• `4,3^3=79.507`;
• `4,4^3=85.184`;
• `4.5^3 = 91.125`;
• `4.6^3 = 97.336`;
• `4.7^3 = 103.823`.

There we go. The third root of `99` is surely between `4.6` and `4.7`.

Just to confirm our calculations, the calculator gave me back `4.626065009182741793092...`, so the answer is correct.