# rational numbers

• Sep 19th 2012, 02:25 AM
Petrus
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 https://webwork.math.su.se/webwork2_...eeb88ed941.png of the form https://webwork.math.su.se/webwork2_...ca422f7991.png, where K and L are rational numbers.
• Sep 19th 2012, 02:40 AM
MarkFL
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?
• Sep 19th 2012, 02:44 AM
Petrus
Re: rational numbers
what do that mean, i dont get it what u mean
• Sep 19th 2012, 02:55 AM
Petrus
Re: rational numbers
u mean multiply 5sqrt(2)-2sqrt(3)?
• Sep 19th 2012, 02:57 AM
MarkFL
Re: rational numbers
You want to multiply the expression by 1 in the form of the conjugate of the denominator over itself, i.e.:

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

Now, perform the multiplication...
• Sep 19th 2012, 03:14 AM
Petrus
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?))
• Sep 19th 2012, 03:20 AM
AtlasSniperman
Re: rational numbers
Quote:

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?))

$\frac{{5\sqrt{2}-2\sqrt{3}+5\sqrt{3}^2}}{{5\sqrt{2}^2-2\sqrt{3}^2}}$

Quote:

[TEY]\frac{{5\sqrt{2}-2\sqrt{3}+5\sqrt{3}^2}}{{5\sqrt{2}^2-2\sqrt{3}^2}}[/TEY]
• Sep 19th 2012, 03:25 AM
MarkFL
Re: rational numbers
$\frac{5\sqrt{2}-2\sqrt{3}}{5\sqrt{2}+2\sqrt{3}}\cdot\frac{5\sqrt{2 }-2\sqrt{3}}{5\sqrt{2}-2\sqrt{3}}=$

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

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

Can you finish?

As far as using $\LaTeX$, 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.
• Sep 19th 2012, 03:54 AM
Petrus
Re: rational numbers
ehmm i cant see how i get it l +k..
• Sep 19th 2012, 03:58 AM
MarkFL
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.
• Sep 19th 2012, 03:59 AM
Petrus
Re: rational numbers
actually i cant finish that, brain stop
• Sep 19th 2012, 04:05 AM
MarkFL
Re: rational numbers
$\frac{50-20\sqrt{6}+12}{50-12}=\frac{62-20\sqrt{6}}{38}=\frac{2(31-10\sqrt{6})}{2\cdot19}=\frac{31-10\sqrt{6}}{19}=$

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