Factoring and rationalizing

• Feb 9th 2013, 09:09 PM
skg94
Factoring and rationalizing
1/ sqroot(x) - 1/2 / x-4

I first simplified by multiplying by 2sqroot(x) = 2-sqroot(x) / (x-4) (2sqroot(x)) than i was going to rationalize the numerator by multiplying by 2+ sqroot(x) but that seemed too complicated.

Where did i go wrong?
• Feb 9th 2013, 09:24 PM
earthboy
Re: Factoring and rationalizing
Quote:

Originally Posted by skg94
1/ sqroot(x) - 1/2 / x-4

I first simplified by multiplying by 2sqroot(x) = 2-sqroot(x) / (x-4) (2sqroot(x)) than i was going to rationalize the numerator by multiplying by 2+ sqroot(x) but that seemed too complicated.

Where did i go wrong?

Why dont you type in latex or give proper brackets?? It helps a lot....
is your question:Rationalize and factorize $\frac{1}{\sqrt{x}}-\frac{1}{2(x-4)}$
• Feb 11th 2013, 08:34 PM
skg94
Re: Factoring and rationalizing
I dont know how to, but well i suppose its the same but its (1/rootx - 1/2 ) / (x-4)
• Feb 11th 2013, 09:29 PM
Soroban
Re: Factoring and rationalizing
Hello, skg94!

Quote:

$\text{Rationalize: }\:\dfrac{\frac{1}{\sqrt{x}} - \frac{1}{2}}{x-4}$

I first simplified by multiplying by $\tfrac{2\sqrt{x}}{2\sqrt{x}}$ .and got: . $\frac{2-\sqrt{x}}{2\sqrt{x}\:\!(x-4)}$

Then I was going to rationalize the numerator by multiplying by $2+ \sqrt{x}$
but that seemed too complicated. . But did you try it?

Where did i go wrong? . Nowhere . . . your work is correct!

You would have: / $\frac{2-\sqrt{x}}{2\sqrt{x}\;\!(x-4)}\cdot {\color{blue}\frac{2+\sqrt{x}}{2 + \sqrt{x}}} \;=\;\frac{4-x}{2\sqrt{x}\;\!(x-4)(2+\sqrt{x})} \;=\; \frac{-(x-4)}{2\sqrt{x}\;\!(x-4)(2+\sqrt{x})}$

. . . . . . . . . . . . . . $=\;\frac{-({\color{red}\rlap{/////}}x-4)}{2\sqrt{x}\;\!({\color{red}\rlap{/////}}x-4)(2+\sqrt{x})} \;=\;\frac{-1}{2\sqrt{x}\;\!(2+\sqrt{x})}$