Respuesta :

[tex]\sqrt{45a^5}[/tex]

[scomposition of 45]

[tex]\sqrt{9*5*a^5}[/tex]

[9 = 3^2, must use this notation (not 3*3)]

[tex]\sqrt{3^2*5*a^5}[/tex]

[We appy one of the proprieties of square roots]

[tex]\sqrt{3^2}* \sqrt{5}*\sqrt{a^5}[/tex]

[now we semplify: we must take out as much as possible all the elements under roots]

[to do that, we must divide the esponent of each element with the index of square roots (2)]

so

[tex]\sqrt{3^2}[/tex], 2/2 = 1

[tex]\sqrt{5}[/tex], 1/2 = 0 with 1 of rest

[tex]\sqrt{a^5}[/tex], 5/2 = 2 with 1 rest

[well, after do that, we can take out the elements under tbhe square roots!]

The quotient of each division is the esponent of the element out of the root

The rest of each division is the esponent of the element under the root

so:

[tex]3^{1}[/tex] (quotient = 1, see the first operation) * [tex]\sqrt{5}[/tex] (rest = 1, see the second operation) * [tex]a^{2}[/tex] (quotient = 2, see the third operation) * [tex]\sqrt{a^1}[/tex] (rest = 1, see the third operation)

The final result is:

3 (=3^1) * a² * √5 * √a

3a²√5a

It's more intuitive and easy, but the explanation (necessary) is very long. If you have other questions, ask me here in the comments! Also sorry for my english, not so good!