Posted by Aniya on Sunday, July 13, 2008 at 4:07pm.
(k+3)^2=(2k1)^2=0
Please help me solve this equation or maybe one like it......I keep doing it wrong.
6p^2(p+1)=4(p+1)5p(p+1)
I also keep doing this problem wrong. I have the answer but I don't understand how its done. Please help!!

Math Urgent  Quidditch, Sunday, July 13, 2008 at 4:13pm
Hint.
Each term on both sides has a factor (p+1). Divide all terms on both sides by (p+1). This will simplify it for you. 
Math Urgent  Aniya, Sunday, July 13, 2008 at 4:53pm
Okay...
Now I got 6p^2=45p(p+1)
I got it now....Thanks!!
the answer is (4/3,1,1/2) 
Math Urgent  Quidditch, Sunday, July 13, 2008 at 4:16pm
Also, because of the (p+1) term remember that p=1 will be a solution.

Math Urgent  Aniya, Sunday, July 13, 2008 at 4:54pm
What about the first question??? Please help me with that one...

Math Urgent  Quidditch, Sunday, July 13, 2008 at 5:06pm
I don't think you copied the first question correctly. Did you intend 2 equal signs?

Math Urgent  Aniya, Sunday, July 13, 2008 at 5:28pm
I copied it wrong....
(k+3)^2=(2k1)^2 .....Is the problem. 
Math Urgent  Quidditch, Sunday, July 13, 2008 at 5:45pm
square both sides, collect like terms and solve.
Here's the first step:
(k+3)^2 = (2k1)^2
k^2 + 6k + 9 = 4k^2  4k + 1
Combine all the like terms. You might factor it out, but I would suggest using the quadratic equation.