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

1 Answer
Oct 21, 2015

Answer:

#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.