# Simplify and Rationalize

• May 20th 2009, 11:53 PM
mvho
Simplify and Rationalize
Question: For x>0, simplify the following:
$\displaystyle \sqrt{x^4-4x^2}/x$

Hint: If x<0, what is :
$\displaystyle \sqrt{x^2}$
• May 21st 2009, 01:46 AM
yeongil
Quote:

Originally Posted by mvho
3). Rationalise the denominator, then simplify:m-4/ sq. root m-4

I take the conjugate which is : sq. root m +4, so my denominator ends up being just m-4. Is my top: Sq root m -16?

So:

Sq. Root m-16/m-4?
Thannks!

No, rationalizing the denominator doesn't mean multiplying by a conjugate. You just multiply the numerator & denominator by the denominator itself:
$\displaystyle \frac{m-4}{\sqrt{m-4}}$

$\displaystyle \frac{m-4}{\sqrt{m-4}}*\frac{\sqrt{m-4}}{\sqrt{m-4}}$

$\displaystyle \frac{(m-4)\sqrt{m-4}}{m-4}$

$\displaystyle \sqrt{m-4}$

I didn't do #1 or #2 because I couldn't figure out what your problems were. Can you use the math tags or put in some parentheses?

01
• May 21st 2009, 02:29 AM
Singaporean
Er i am assuming that the square root applies to only $\displaystyle x^4$

1.$\displaystyle \frac{\sqrt (x^4)-(4x^2)}{x}$

$\displaystyle \frac{(x^2)-(4x^2)}{x}$

$\displaystyle x-(4x)$

$\displaystyle -3x$

lol think so
• May 21st 2009, 01:04 PM
mvho
Hmm... The square root for question #3, ends at the m. Ill have to find out how to use math tags.

Any links to learn how to use math tags? I need that sq. root symbol!
• May 21st 2009, 02:07 PM
yeongil
Quote:

Originally Posted by mvho
Hmm... The square root for question #3, ends at the m. Ill have to find out how to use math tags.

$\displaystyle \frac{m - 4}{\sqrt{m} - 4}$?
Then yes, you would multiply by the conjugate:
$\displaystyle \frac{m - 4}{\sqrt{m} - 4} * \frac{\sqrt{m} + 4}{\sqrt{m} + 4}$
$\displaystyle \frac{(m - 4)(\sqrt{m} + 4)}{(\sqrt{m})^{2} - 4^{2}}$
$\displaystyle \frac{(m - 4)(\sqrt{m} + 4)}{m - 16}$
Dunno if it's necessary to FOIL the numerator. There's nothing else to simplify that I can see.

01
• May 21st 2009, 02:28 PM
mvho
yeongil: Thats right. That is what I got too. How do yo use these math tags though? Is there a FAQ somewhere?
Quote:

Er i am assuming that the square root applies to only http://www.mathhelpforum.com/math-he...1fb8a76a-1.gif

The square root applies to the entire numerator. I need to learn about these math tags!!
• May 22nd 2009, 02:38 AM
Singaporean
Quote:

Originally Posted by mvho
yeongil: Thats right. That is what I got too. How do yo use these math tags though? Is there a FAQ somewhere?

The square root applies to the entire numerator. I need to learn about these math tags!!

http://www.mathhelpforum.com/math-he...-tutorial.html

the file attached there should explain everything.
• May 22nd 2009, 01:30 PM
mvho
Thanks for the link! I have revised my question now, so it Should be much clearer.

I am unclear what the hint is implying, does anyone have a clue?

$\displaystyle -\sqrt{x^2-4}$