Square root within a square root

[Ross MC]

Hoping someone can push me in the right direction with this one. Plume snookered.

It's to simplify:
2√3(3+√3)

Guessing first calculate (a^2 - b^2*c) in the square, though the 2 is throwing this an I'm not sure how the answer is 6√3 + 6, an not 18√3 ?

 

[coolul007]

Hoping someone can push me in the right direction with this one. Plume snookered.

It's to simplify:
2√3(3+√3)

Guessing first calculate (a^2 - b^2*c) in the square, though the 2 is throwing this an I'm not sure how the answer is 6√3 + 6, an not 18√3 ?

It's just the Distributive property:
$(2)(\sqrt{3})(3 + \sqrt{3})$
$(2)(\sqrt{3})(3 ) + (2)(\sqrt{3})(\sqrt{3})$
$(2)(3 )(\sqrt{3}) + (2)(\sqrt{3})(\sqrt{3})$

 

[Ross MC]

Thanks coolul007, haven't really been shown the Distributive way of working with squares like that, has helped shed some light on another tricky simplication.

 

[haruspex]

The title said 'square root within a square root', which implies you meant 2√(3(3+√3)), but the answer you say is correct matches (2√3)(3+√3).

 

[Mentallic]

There is no square root within a square root as your title suggested

Unless you meant that 2√3(3+√3) as $$2\sqrt{3(3+\sqrt{3})}$$ but that doesn't follow from the required answer, so yep, this problem can simply be solved with the distributive property.

edit: beaten to it.

 

[Mark44]

Mod note: Moving this thread to the Precalculus section under Homework & Coursework, which is where the OP should have started this thread.