SOLUTION: Hi Tutors, Can someone see if I did this problem right and if not can you tell me what I did wrong. Solve the equations. sqrt 4x + 1 + 3 =0 sqrt 4x + 1 = -3 sqrt 4x + 1

Algebra ->  Radicals -> SOLUTION: Hi Tutors, Can someone see if I did this problem right and if not can you tell me what I did wrong. Solve the equations. sqrt 4x + 1 + 3 =0 sqrt 4x + 1 = -3 sqrt 4x + 1      Log On


   

Question 119883: Hi Tutors, Can someone see if I did this problem right and if not can you tell me what I did wrong. Solve the equations.

sqrt 4x + 1 + 3 =0
sqrt 4x + 1 = -3
sqrt 4x + 1^2 =(-3)^2
4x + 1= 9
4x = 9 – 1 Subtract
4x = 8 Divide by 2
x= 2 is the answer
How can I check to see if this is the right answer.


One more thing I am stuck on this problem because of the x can someone please help.
sqrt 2x + 1 + x = 7
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%284x+%2B+1%29+%2B+3+=+0
sqrt%28+4x+%2B+1%29+=+-3
%28sqrt%284x+%2B+1%29%29%5E2+=%28-3%29%5E2
4x+%2B+1=+9
4x+=+9+%96+1 Subtract
4x+=+8 Divide by 2
x+=+2 is the answer
How can I check to see if this is the right answer.
You can plug this value of x back into the original equation,
but first notice what you got above:
%28sqrt%284x+%2B+1%29%29%5E2+=%28-3%29%5E2
This is telling you that what you're squaring on the right is
a negative number, so The square root on the left must be the
negative square root, which, when squared gives a positive
Going back to:
sqrt%284x+%2B+1%29+%2B+3+=+0
sqrt%284%2A2+%2B+1%29+%2B+3+=+0
sqrt%289%29+=+-3
This is true, one of the square roots of 9 is -3.
-3+=+-3
OK