Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

Login

Username

Password

Not registered yet?

#1 2006-05-05 17:57:25

4littlepiggiesmom
Member

Offline

Solving Radical Equations

1. 8/(x+4) +1 = 5x/(x^2-2x-24)

2. 2x(x+3) -  x/(x+7) = (x^2-1)/(x^2+10x+21)

Any one able to show me where to start after figuring out the LDC would be my hero for the day!

The LDC for 1 is (x-6)(x+4)
and far 2 it's (x+3)(x+7) correct?

#2 2006-05-05 18:48:18

4littlepiggiesmom
Member

Offline

Re: Solving Radical Equations

First ones answer is x= 8 and -9...corret ??? but the other one I can get anywher with is at alk..
I get this
(2x)(x+7) - (x) (3+3) = (y^2 -1)
then I get (2x^2 +14 x)(x-3x) = (y^2 -1) but then I keep getting zero on both side???? Help please!!!

#3 2006-05-05 22:45:16

George,Y
Super Member

Offline

Re: Solving Radical Equations

2x(x+7)-x(x+3) = x-1
x(2x+14-x-3)=x-1
11x=-1
x=-1/11


X'(y-Xβ)=0

Board footer

Powered by FluxBB