Here's an image of my problem.

Printable View

- Apr 24th 2006, 07:41 AMcoopsterdudeHelp me solve the square root of a square root.
Here's an image of my problem.

- Apr 24th 2006, 08:03 AMearbothQuote:

Originally Posted by**coopsterdude**

there are a few different ways to simplify this term:

$\displaystyle \sqrt{12 \cdot \sqrt{3}}=\sqrt{ \sqrt{12^2 \cdot3}}= \sqrt[4]{432}$

or:

$\displaystyle \sqrt{12 \cdot \sqrt{3}}=\sqrt{2^2 \cdot 3 \cdot \sqrt{\cdot3}}= 2 \cdot \sqrt{\sqrt{3^2 \cdot 3}}=2 \cdot \sqrt[4]{27} $

Greetings

EB - Apr 29th 2006, 01:13 AMrgep
... or $\displaystyle \sqrt{12\sqrt{3}} = (12(3^{1/2}))^{1/2} = (12)^{1/2}((3)^{1/2})^{1/2}$ = $\displaystyle (2^2.3)^{1/2} 3^{1/4} = (2^2)^{1/2} 3^{1/2+1/4} = 2.3^{3/4} = 2\sqrt[4]{27}$.