# Simplifying, its easy.

• September 26th 2007, 01:30 PM
Jeavus
Simplifying, its easy.
(100 / 4 root x) - (36 /x^2)

I get this far:

(10^2 / x^(1/2)) - (6^2 /x^2)

EDIT: Anyone have any idea?
• September 26th 2007, 01:44 PM
topsquark
Quote:

Originally Posted by Jeavus
(100 / 4 root x) - (36 /x^2)

I get this far:

(10^2 / x^(1/2)) - (6^2 /x^2)

(You missed a 4 in the first fraction, by the way.)

You have two options as I see it.

Find a common denominator. The LCM of $4x^{1/2} \text{ and }x^2$ is $4x^2$.

So
$\frac{100}{x^{1/2}} \cdot \frac{x^{3/2}}{x^{3/2}} - \frac{36}{x^2} \cdot \frac{4}{4}$

$= \frac{100x^{3/2} - 144}{4x^2} = \frac{25x^{3/2} - 72}{x^2}$

2. Simplify each term independently.
$\frac{100}{4\sqrt{x}} - \frac{36}{x^2}$

$= \frac{25}{\sqrt{x}} - \frac{36}{x^2}$

$= \frac{25 \sqrt{x}}{x} - \frac{36}{x^2}$

Now factor:
$= \frac{1}{x} \cdot \left ( 25 - \frac{36}{x} \right )$

-Dan
• September 26th 2007, 01:50 PM
Jeavus
I think you misinterpreted the question.

4 root x

It looks like this:

http://img255.imageshack.us/img255/8985/4rootuh6.png

http://img255.imageshack.us/img255/8985/4rootuh6.png
• September 26th 2007, 01:54 PM
topsquark
Quote:

Originally Posted by Jeavus
I think you misinterpreted the question.

4 root x

It looks like this:

http://img255.imageshack.us/img255/8985/4rootuh6.png

http://img255.imageshack.us/img255/8985/4rootuh6.png

Ahhhhhh....

Hmmmm.... It doesn't seem like we'd want to add the fractions here, so I would simplify the first term, then factor:
$\frac{100}{\sqrt[4]{x}} - \frac{36}{x^2}$

$= \frac{100}{\sqrt[4]{x}} \cdot \frac{\sqrt[4]{x^3}}{\sqrt[4]{x^3}} - \frac{36}{x^2}$

$= \frac{100 \sqrt[4]{x^3}}{x} - \frac{36}{x^2}$

$= \frac{1}{x} \left ( 100 \sqrt[4]{x^3} - \frac{36}{x} \right )$

-Dan
• September 26th 2007, 01:56 PM
topsquark
I suppose you could also treat this as the difference between two squares:
$\frac{100}{x^{1/4}} - \frac{36}{x^2}$

$= \left ( \frac{10}{x^{1/8}} + \frac{6}{x} \right )\left ( \frac{10}{x^{1/8}} - \frac{6}{x} \right )$

$= 4 \left ( \frac{5}{x^{1/8}} + \frac{3}{x} \right )\left ( \frac{5}{x^{1/8}} - \frac{3}{x} \right )$

-Dan
• September 26th 2007, 02:11 PM
Jeavus
How does it become 10 / x^1/8?
• September 26th 2007, 04:11 PM
topsquark
Quote:

Originally Posted by Jeavus
How does it become 10 / x^1/8?

Recall that $a^2 - b^2 = (a + b)(a - b)$.

So your question becomes: How do you take the square root of $\frac{100}{x^{1/4}}$?

I assume you can do the square root of 100, so how do you do the square root of $x^{1/4}$?
$\sqrt{x^{1/4}} = \left ( x^{1/4} \right ) ^{1/2} = x^{1/4 \cdot 1/2} = x^{1/8}$

Thus
$\sqrt{\frac{100}{x^{1/4}}} = \frac{10}{x^{1/8}}$

-Dan