# 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 ?

Gold Member
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, havent really been shown the Distributive way of working with squares like that, has helped shed some light on another tricky simplication.

Homework Helper
Gold Member
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).

Homework Helper
There is no square root within a square root as your title suggested :tongue:

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.

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