Question

How do you simplify `\sqrt { 29x ^ { 6} y ^ { 15} }`?

Answer 1

`x^3y^7sqrt(29y)`

Explanation:

The only part of the radicand (the term under the square root) which has an exact square root is `x^6`

Write the radicand as factors, using squares where possible:
Find the roots of any perfect squares.

`sqrt(29 xx color(red)(x^6 xx y^14) xx y) = color(red)(x^3y^7)sqrt(29y)`

Prime numbers are never perfect squares.

Any base with an even index is a perfect square.

Answer 2

`x^3` `y^7` `sqrt(29y)`

Explanation:

Since we're trying to find the square root, try writing the variables to the power of two so it can be factored out easily.

So the first step will be
`sqrt(29x^6y^14y)`

You can further break this down or simply pull out the roots. Let's break it down again...

`sqrt(29x^2x^2x^2y^2y^2y^2y^2y^2y^2y^2y)`

The above is just to show how the roots are taken out.

29 has no square root so that stays in. The square root of `x^2` is x. So since there are three of that, we get `x^3` out. Same for `y^2`, there are 7 of them so that comes out leaving `y^1` in.

At the end we have;

`x^3` `y^7` `sqrt(29y)`