How to rationalize the numberator?

[dtcool2003]

Here is the question: the square root of X minus 3, but the 3 is not under the square root divided by x-9? also another question: the square root of X minus 2, but the 2 is not under the square root divided by 4-x? anyone can help me?

 

[dtcool2003]

How to rationalize the numerator?

Here is the question: sqrt(x) - 3/ x-9 ? also another question: sqrt(x) -2/ 4-x ? anyone can help me? Thanks

 

[daveb]

So you mean sqrt(x)/(x-9) - 3/(x-9)?

 

[dtcool2003]

No

no I mean, sqrt(x) - 3/ x-9 ? also another question: sqrt(x) -2/ 4-x ? anyone can help me? Thanks and I need to rationalize the numerator

 

[dextercioby]

Nope, i guess he means

$$\frac{\sqrt{x} -3}{x-9}$$ and $$\frac{\sqrt{x} -2}{4-x}$$

When rationalizing, remember that $a^2 - b^2 \equiv (a+b)(a-b)$. So take [/itex] a=\sqrt{x} [/itex]. What is 'b' equal to ?

 

[dtcool2003]

yeah

those are the right equations u just put up, but i don't know how to rationalize those equations?

 

[dtcool2003]

yeah

I got x-9/ x-9sqrt(x) + 3x - 27 is that right?

 

[HallsofIvy]

$$\frac{\sqrt{x}-3}{x^2- 9}$$
That it?

Use (a- b)(a+ b)= a²- b²: multiply both numerator and denominator by $\sqrt{3}+ 3$ and the squareroot will disappear from the numerator. It will, of course, show up in the denominator.

For the second, assuming you have a fraction with $\sqrt{x}- 2$ in either numerator or denominator, multiply both numerator and denominator by $\sqrt{x}+ 2$

 

[HallsofIvy]

This is not a "Linear and Abstract Algebra" question and it was also posted in the "General Math" section so I am merging the two threads in that section.

 

[ActionPotential]

$$\frac {\sqrt{x}-3}{x-9}$$

Multiply:
Both $${x-9}$$ and $${\sqrt{x}-3}$$

By the conjugate of $${\sqrt{x}-3}$$ which is $${\sqrt{x}+3}$$.

$$\frac {\sqrt{x}-3( \sqrt{x}+3)} {x-9( \sqrt{x}+3)}$$

From here, you should be able to do some algebra, to rationalize the numerator. Do you know what a conjugate is, and why it is used? That might be your issue with not understanding it.