Question: For x>0, simplify the following:
$\displaystyle
\sqrt{x^4-4x^2}/x
$
Hint: If x<0, what is :
$\displaystyle
\sqrt{x^2}
$
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
Please do! So you mean
$\displaystyle \frac{m - 4}{\sqrt{m} - 4}$?
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
http://www.mathhelpforum.com/math-he...-tutorial.html
the file attached there should explain everything.
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?
Here is my answer:
$\displaystyle
-\sqrt{x^2-4}
$
I know I can simplify further (difference of squares) but this looks neater and I m done fooling around with latex for 1 day