Simplify and Solve Radical and Square Equations - Homework Help

[littlefirefly]

Homework Statement

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

Homework 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.

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.

 

[littlefirefly]

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

 

[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.

 

[cristo]

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

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.

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

 

[Gib Z]

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

 

[littlefirefly]

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

 

[cristo]

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.

 

[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.

 

[Gib Z]

It doesn't look like that factors..

 

[drpizza]

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

 

[Gib Z]

Ahh god damn I am an idiot..good idea