1. Jan 21, 2007

littlefirefly

1. The problem statement, all variables and given/known data

I need the first equation simplified, the second equation solved, and the third factored. I've been trying to solve these three forever today.

2. Relevant equations
1) 6+√3 ÷ 2-2√3 (those are radical signs)
2) √k+9-√k=√3 (the k+9 is all under the radical sign)
3) (2k+3)2-(2x+3)(y-2)-20(y-2)2 (the "2" after (2k+3) ans (y-2) is suppose to be a square sign.

3. The attempt at a solution
I thought for number 1) I should use the conjugent (sp) but I wasn't sure if I foil or not and I got three different answers that don't look right. Number 2) I have no idea, I tired to make the √k by itself on one side that then square everything, but when I plugged the number I got back in the equation it didn't work so I don't know what I'm doing. Help would be MUCH appreciated.

2. Jan 21, 2007

littlefirefly

I didn't put the equations in the right spot, i'm sorrrry....

3. Jan 21, 2007

Gib Z

Welcome to Physicsforums!

Umm, next time post the title to be like, Simple Equations or something. When people see please help theyre less inclined to help XD

Ok.

The first equation, I can't seem to understand what you mean...sorry about that

The second one, Square both sides :D Then we get (k+9) - 2 √(k^2 + 9k) - (k)= 3
K's cancel out, minus 3 from both sides, take the -2√(k^2 + 9k) to the other side, cancel common factors

We end up with 3=√(k^2 - 9k), square both sides again.
9=k^2-9k, or k^2-9k-9=0. Thats a simple quadratic equation. I hope you know the quadratic formula!

3) I Think all that can go down to is $$(3-2x-20 (2)^{\frac{1}{2}})(y-2)+ 2^{\frac{1}{2}} (2k+3)$$ where the 2 to the power of halfs are square root 2.

Sorry I couldn't be of more help, Good Luck.

4. Jan 21, 2007

cristo

Staff Emeritus
For question 1, multiply top and bottom by $2+2\sqrt3$ and expand the brackets.

I'm not sure how this can be correct as the original expression has a k2 in, and yours doesn't.

5. Jan 21, 2007

Gib Z

!!>.< Sorry I thought the 2s after the things were root 2, not to the pwer of 2 >.< Shoot me....

6. Jan 21, 2007

littlefirefly

thanks for the help, I just couldn't figure those questions out =)

7. Jan 21, 2007

cristo

Staff Emeritus
For 3, I'm afraid if you can't see anything that will easily factor, you'll have to expand it all, then try and factor.

8. Jan 21, 2007

drpizza

I'm guessing on the 3rd one you have, you didn't really mean to have both x and k in there. Here's a neat trick:
"(2k+3)2-(2x+3)(y-2)-20(y-2)2 (the "2" after (2k+3) ans (y-2) is suppose to be a square sign"
I'm going to assume the x is really supposed to be a k.
Temporarily, let's let A=(2k+3) and B=(y-2)
Then, you have A^2 -AB - 20B^2
Can you factor that into two binomials?
If you can, then after you factor it, substitute 2k+3 back in for A and y-2 back in for B.

9. Jan 22, 2007

Gib Z

It doesnt look like that factors..

10. Jan 22, 2007

drpizza

A^2-AB-20B^2 is
(A-5B)(A+4B)
It does factor. Now substitute back in for A and B.

11. Jan 22, 2007

Gib Z

Ahh god damn Im an idiot..good idea