Question

Question #69536

Answer 1

We get, `5 xy^3sqrt(6xz)` from `sqrt(150x^3y^6z)`

Explanation:

`5xy^3sqrt(6xz)` from `sqrt(150x^3y^6z)`

Consider
`sqrt(150x^3y^6z)`

As `a^ mxxa^n = a^(m+n)`,

`=> sqrt((25xx6)xx(x^2xxx)xx(y^2xxy^4)xxz)`

`=> sqrt((5^2xx6)xx(x^2xxx )xx(y^2xxy^2xxy^2)xxz)`

Taking the roots of the square terms outside the radical sign:

`=> 5xx(x)xxyxxyxxyxx sqrt(6xz)`

`=>5 xy^3sqrt(6xz)`

Answer 2

`5xy^3sqrt(6xz)`

Explanation:

`color(green)("The trick is to look for squared values")`

`color(blue)("Dealing with the number part")`

We have 150.

Just for a moment forget that this is hundreds and just focus on the 15 part. It is known that `3xx5=15`

Now we deal with the hundreds part. It is known that 15xx10=150. So putting all of this together we have: `3xx5xx10=150`

We know that `2xx5=10`. Again we can substitute this back in giving: `3xx5xx2xx5=150`

Notice that within this we have `5xx5` so we can write:

`3xx2xx5^2`

Bothe 3 and 2 are prime numbers so there is no way we can square root them. However, `sqrt(5^2)=5` so we can square root that part.

`color(red)(sqrt(150)=5sqrt(3xx2)=5sqrt(6))`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Dealing with the variable part (letters)")`

`x^3y^6z`

`x^3` is the same as `x^2xx x`
`y^6` is the same as `y^(2+2+2)=y^2xxy^2xxy^2`

A single `z` we can do nothing with.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Putting it all together")`

So we have: `5sqrt(6xxx^2 xx x xxy^2xxy^2xxy^2xxz)`

Taking all the squared values out of the root

`5xy^3sqrt(6xz)`