# Thread: rational numbers

1. ## rational numbers

hi i had this problem on my test and did not find any way to solve it but im verry intressted how to solve it! plz any1 can help:O?
Write the expression of the form , where K and L are rational numbers.

2. ## Re: rational numbers

What you want to do here is rationalize the denominator, then simplify, and you will get the required form.

What would your first step be to rationalize the denominator?

3. ## Re: rational numbers

what do that mean, i dont get it what u mean

4. ## Re: rational numbers

u mean multiply 5sqrt(2)-2sqrt(3)?

5. ## Re: rational numbers

You want to multiply the expression by 1 in the form of the conjugate of the denominator over itself, i.e.:

$\displaystyle \frac{5\sqrt{2}-2\sqrt{3}}{5\sqrt{2}+2\sqrt{3}}\cdot\frac{5\sqrt{2 }-2\sqrt{3}}{5\sqrt{2}-2\sqrt{3}}$

Now, perform the multiplication...

6. ## Re: rational numbers

(5sqrt(2)^2-2(5sqrt(2)*2sqrt(3))+5sqrt(3)^2) / (5sqrt(2)^2+2-sqrt(3)^2 )

(how do i get it in code like u ( latex?))

7. ## Re: rational numbers

Originally Posted by Petrus
(5sqrt(2)^2-2(5sqrt(2)*2sqrt(3))+5sqrt(3)^2) / (5sqrt(2)^2+2-sqrt(3)^2 )

(how do i get it in code like u ( latex?))
$\displaystyle \frac{{5\sqrt{2}-2\sqrt{3}+5\sqrt{3}^2}}{{5\sqrt{2}^2-2\sqrt{3}^2}}$

[TEY]\frac{{5\sqrt{2}-2\sqrt{3}+5\sqrt{3}^2}}{{5\sqrt{2}^2-2\sqrt{3}^2}}[/TEY]
with TEX instead of TEY

8. ## Re: rational numbers

$\displaystyle \frac{5\sqrt{2}-2\sqrt{3}}{5\sqrt{2}+2\sqrt{3}}\cdot\frac{5\sqrt{2 }-2\sqrt{3}}{5\sqrt{2}-2\sqrt{3}}=$

$\displaystyle \frac{5^2\cdot2-2\cdot2\cdot5\cdot\sqrt{2}\cdot\sqrt{3}+2^2\cdot3} {5^2\cdot2-2^2\cdot3}=$

$\displaystyle \frac{50-20\sqrt{6}+12}{50-12}$

Can you finish?

As far as using $\displaystyle \LaTeX$, there are tutorials here and online that will do a much better job of explaining its usage than I can do in a single post.

9. ## Re: rational numbers

ehmm i cant see how i get it l +k..

10. ## Re: rational numbers

No, you can only cancel or divide out factors, not addends. Add/subtract like terms (integers) in both the numerator and denominator, then reduce by factoring.

11. ## Re: rational numbers

actually i cant finish that, brain stop

12. ## Re: rational numbers

$\displaystyle \frac{50-20\sqrt{6}+12}{50-12}=\frac{62-20\sqrt{6}}{38}=\frac{2(31-10\sqrt{6})}{2\cdot19}=\frac{31-10\sqrt{6}}{19}=$

$\displaystyle \left(\frac{31}{19} \right)+\left(-\frac{10}{19} \right)\sqrt{6}$