Schrodinger Group Title I'm having trouble finding out how I can move from 4/(sqrt(2)-sqrt(6)) to -sqrt(2) - sqrt(6). I really can't figure this out for the life of me, and I have no idea why. 2 years ago 2 years ago

1. hartnn

hint :multiply and divide by (sqrt(2)+sqrt(6)

2. Schrodinger

I tried that, but i'm getting just a batsh*t crazy mess when I try to do that. I'm assuming you're talking about multiplying the numerator and denominator by the numerator? I must be doing something wrong there.

3. hartnn

not with numerator, with denominator, but with middle sign reversed , (sqrt(2)+sqrt(6)

4. Schrodinger

Okay, that's cool, but how do you just know to do that?

5. hartnn

its called rationalizing the denominator, if there is $$\large a+\sqrt b$$ in the denominator, u multiply and divide by $$\large a-\sqrt b$$

6. hartnn

similar case when u want to rationalize the numerator.....

7. Schrodinger

Okay, cool. I've actually never seen that done before, at least formally, before. Thanks!

8. hartnn

welcome ^_^

9. hartnn

try it out here, and tell me if u don't get it....

10. Schrodinger

$\frac{4}{\sqrt{2}-\sqrt{6}}$... $\frac{ 4 }{\sqrt{2}-\sqrt{6}} * \frac{ \sqrt{2}+\sqrt{6} }{ \sqrt{2}+\sqrt{6} }$... $\frac{ 4\sqrt{2}+4\sqrt{6} }{ 2+\sqrt{12}-\sqrt{12}-6 }$... $\frac{ 4\sqrt{2}+4\sqrt{6} }{ 2 - 6 }$... $\frac{ 4\sqrt{2}+4\sqrt{6} }{ -4 }$... $-\sqrt{2}-\sqrt{6}$ Correct?

11. hartnn

absolutely ! good work :)

12. Schrodinger

Thanks. I'm pretty worried/bothered at the fact that I was never formally taught that concept in middle or high school. My teachers were the type to kind of go, "Oh well you just put this here and do that and then hooray!" without giving a concrete or logically consistent approach to how to simplify or solve certain things. It's like it was assumed that I understood. Man, public school isn't exactly wonderful.

13. hartnn

don't worry, whatever u can't understand, ask here, we'll try our best to make u understand :)