# Simplifying This Expression

• Aug 14th 2011, 03:24 PM
YoungMarbleGiant
Simplifying This Expression
I have the following:

${(4 - \frac{1}{(x+1)^{2}})^{1/2}}$

How do you go about simplifying this? DO you simplify each part individually by the 1/2 exponent? Thank you for any tips.
• Aug 14th 2011, 03:36 PM
anonimnystefy
Re: Simplifying This Expression
you can't do it individually because it is the same as:
$\sqrt{4-\frac{1}{(x+1)^2}}$
$\sqrt{a-b}\neq\sqrt{a}-\sqrt{b}$
to do this you have to subtract $4-\frac{1}{(x-1)^2}$.can you do that?
• Aug 14th 2011, 03:52 PM
YoungMarbleGiant
Re: Simplifying This Expression
Quote:

Originally Posted by anonimnystefy
you can't do it individually because it is the same as:
$\sqrt{4-\frac{1}{(x+1)^2}}$
$\sqrt{a-b}\neq\sqrt{a}-\sqrt{b}$
to do this you have to subtract $4-\frac{1}{(x-1)^2}$.can you do that?

To get $\frac{4(x-1)^{2} - 1}{(x-1)^{2}}$ ... ?

If that's correct, am I then able to take each term to the 1/2 exponent because it's all expressed with the same denominator?
• Aug 14th 2011, 05:12 PM
Prove It
Re: Simplifying This Expression
Quote:

Originally Posted by YoungMarbleGiant
To get $\frac{4(x-1)^{2} - 1}{(x-1)^{2}}$ ... ?

If that's correct, am I then able to take each term to the 1/2 exponent because it's all expressed with the same denominator?

As you were told above, the square root of a sum/difference is NOT equal to the sum/difference of the square roots. So you can't take each term in the numerator to the 1/2 power.

However, the square root of a product/quotient DOES equal the product/quotient of the square roots. So you CAN take the square root of the numerator and the square root of the denominator.
• Aug 14th 2011, 05:42 PM
YoungMarbleGiant
Re: Simplifying This Expression
Quote:

Originally Posted by Prove It
As you were told above, the square root of a sum/difference is NOT equal to the sum/difference of the square roots. So you can't take each term in the numerator to the 1/2 power.

However, the square root of a product/quotient DOES equal the product/quotient of the square roots. So you CAN take the square root of the numerator and the square root of the denominator.

So it simplifies to the square root of the whole numerator, over (x-1)... is that correct? Sorry if I've not followed you here.
• Aug 14th 2011, 06:50 PM
Prove It
Re: Simplifying This Expression
Quote:

Originally Posted by YoungMarbleGiant
So it simplifies to the square root of the whole numerator, over (x-1)... is that correct? Sorry if I've not followed you here.

The denominator should be |x - 1|, otherwise you're correct.