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Square root within a square root

  • Thread starter Ross MC
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  • #1
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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 ?
 

Answers and Replies

  • #2
coolul007
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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:
[itex](2)(\sqrt{3})(3 + \sqrt{3})[/itex]
[itex](2)(\sqrt{3})(3 ) + (2)(\sqrt{3})(\sqrt{3})[/itex]
[itex](2)(3 )(\sqrt{3}) + (2)(\sqrt{3})(\sqrt{3})[/itex]
 
  • #3
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Thanks coolul007, havent really been shown the Distributive way of working with squares like that, has helped shed some light on another tricky simplication.
 
  • #4
haruspex
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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).
 
  • #5
Mentallic
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There is no square root within a square root as your title suggested :tongue:

Unless you meant that 2√3(3+√3) as [tex]2\sqrt{3(3+\sqrt{3})}[/tex] 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.
 
  • #6
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Mod note: Moving this thread to the Precalculus section under Homework & Coursework, which is where the OP should have started this thread.
 

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