# simplying expressions (indices)

• Sep 9th 2008, 08:40 AM
jvignacio
simplying expressions (indices)
$\displaystyle 4x = \frac{1}{\sqrt{3}}$

any help with this one? thanksss you
• Sep 9th 2008, 08:48 AM
Simplicity
Quote:

Originally Posted by jvignacio
$\displaystyle 4x = \frac{1}{\sqrt{3}}$

any help with this one? thanksss you

I assume you want to make $\displaystyle x$ the subject and leave it in simplified form?

$\displaystyle 4x = \frac{1}{\sqrt{3}}$

Divide by $\displaystyle 4$

$\displaystyle x = \frac{1}{4\sqrt{3}}$

You cannot have a fraction with a surd on the denominator so change it's value to get the surd into the numerator. To do this, multiply the denominator and the denominator by $\displaystyle \sqrt{3}$.

$\displaystyle \frac{1}{4\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{(4)(\sqrt{3})(\sqrt{3})} = \frac{\sqrt{3}}{(4)(3)} = \frac{\sqrt{3}}{12}$

$\displaystyle \therefore x = \frac{\sqrt{3}}{12}$
• Sep 9th 2008, 08:55 AM
jvignacio
Quote:

Originally Posted by Air
I assume you want to make $\displaystyle x$ the subject and leave it in simplified form?

$\displaystyle 4x = \frac{1}{\sqrt{3}}$

Divide by $\displaystyle 4$

$\displaystyle x = \frac{1}{4\sqrt{3}}$

You cannot have a fraction with a surd on the denominator so change it's value to get the surd into the numerator. To do this, multiply the denominator and the denominator by $\displaystyle \sqrt{3}$.

$\displaystyle \frac{1}{4\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{(4)(\sqrt{3})(\sqrt{3})} = \frac{\sqrt{3}}{(4)(3)} = \frac{\sqrt{3}}{12}$

$\displaystyle \therefore x = \frac{\sqrt{3}}{12}$

thanks mate! but I dont need to solve for x, just simplify the expression
• Sep 9th 2008, 08:59 AM
Simplicity
Quote:

Originally Posted by jvignacio
thanks mate! but I dont need to solve for x, just simplify the expression

If the question is $\displaystyle 4x = \frac{1}{\sqrt{3}}$, like you said, and all it says is 'simplify the expression' then I would have leave it simplified as:

$\displaystyle x = \frac{\sqrt{3}}{12}$

EDIT: For your reference, $\displaystyle \frac{1}{\sqrt{x}} = x^{-\frac12}$
• Sep 9th 2008, 09:04 AM
jvignacio
Quote:

Originally Posted by Air
If the question is $\displaystyle 4x = \frac{1}{\sqrt{3}}$, like you said, and all it says is 'simplify the expression' then I would have leave it simplified as:

$\displaystyle x = \frac{\sqrt{3}}{12}$

EDIT: For your reference, $\displaystyle \frac{1}{\sqrt{x}} = x^{-\frac12}$

yeah yeah i knew that.
okay mate thanks alot ! :)