SOLUTION: COULD SOMEONE PLEASE SHOW ME HOW TO SOLVE THIS EQUAATION? i CAN ONLY GET SO FAR. {{{(SQRT 4x+1)+3=0}}} This is what i have gotten cant seem to remember how to get any further.

Algebra ->  Radicals -> SOLUTION: COULD SOMEONE PLEASE SHOW ME HOW TO SOLVE THIS EQUAATION? i CAN ONLY GET SO FAR. {{{(SQRT 4x+1)+3=0}}} This is what i have gotten cant seem to remember how to get any further.       Log On


   

Question 98760: COULD SOMEONE PLEASE SHOW ME HOW TO SOLVE THIS EQUAATION? i CAN ONLY GET SO FAR.
%28SQRT+4x%2B1%29%2B3=0
This is what i have gotten cant seem to remember how to get any further.
%28sqrt4x%2B1%29=0-3
%28sqrt4x%2B1%29%5E2=-3%5E2
4x%2B1=9
4x=9-1
4x=1

Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
%28SQRT+4x%2B1%29%2B3=0
%28sqrt4x%2B1%29=-3
%28sqrt4x%2B1%29%5E2=%28-3%29%5E2
4x%2B1=9
4x=9-1
4x=8you subtracted 1 from 9 and got 1
divide 4 into each side: x=2
good work! small mathematical mistake.
Check:
sqrt((4*2)+1)=+3 and -3 (+3 is extraneous for this equation)
Ed