text: # solving equations w/ radicals and exponents problem, stuck
  
  • Aug 5th 2011, 01:03 PM
  mathmathmathmathmathmathm
  solving equations w/ radicals and exponents problem, stuck
  -6 - 2x = 4 times sqrt of x + 6
  
  this was a part of a larger problem but i see this is where it screws up, the answer is -5 as the solution set
  
  what ive been doin w/ problems before this one was divide that 4 in both places and it always fit nicely but not in this problem, iam really frustrated can someone gimmie a hint? thanks a bunch
  • Aug 5th 2011, 01:04 PM
  mathmathmathmathmathmathm
  Re: solving equations w/ radicals and exponents problem, stuck
  heres the orginal problem
  
  sqrt of 4 - x - sqrt of x + 6 = 2
  • Aug 5th 2011, 01:07 PM
  e^(i*pi)
  Re: solving equations w/ radicals and exponents problem, stuck
  Quote:
  
  Originally Posted by mathmathmathmathmathmathm
  -6 - 2x = 4 times sqrt of x + 6
  
  this was a part of a larger problem but i see this is where it screws up, the answer is -5 as the solution set
  
  what ive been doin w/ problems before this one was divide that 4 in both places and it always fit nicely but not in this problem, iam really frustrated can someone gimmie a hint? thanks a bunch
  
  You can square both sides:
  
  $(-6-2x)^2 = (4\sqrt{x+6})^2 \ \Leftrightarrow \ 4x^2+24x+36 = 16(x+6)$
  
  Solve for x and check for extraneous solutions using the original equation. Note that due to the domain $x \geq -6$
  
  EDIT: Your equation in the OP is fine but there is an additional domain restriction $-6 \leq x \leq 4$
  
  I've put the working to show this in a spoiler to keep the post more coherent.
  
  Spoiler:
  Quote:
  
  Originally Posted by mathmathmathmathmathmathm
  heres the orginal problem
  
  $\sqrt{4 - x} - \sqrt{x + 6} = 2$
  
  $\sqrt{4-x} = 2 + \sqrt{x+6}$
  
  $4-x = 4 + 4\sqrt{x+6} + x+6 = 10 + x + 4\sqrt{x+6}$
  
  If you tidy that up a bit: $-6-2x = 2\sqrt{x+6}$
  • Aug 5th 2011, 07:45 PM
  Prove It
  Re: solving equations w/ radicals and exponents problem, stuck
  Quote:
  
  Originally Posted by mathmathmathmathmathmathm
  -6 - 2x = 4 times sqrt of x + 6
  
  this was a part of a larger problem but i see this is where it screws up, the answer is -5 as the solution set
  
  what ive been doin w/ problems before this one was divide that 4 in both places and it always fit nicely but not in this problem, iam really frustrated can someone gimmie a hint? thanks a bunch
  
  Is your equation $\displaystyle -6-2x = 4\sqrt{x} + 6$ or $\displaystyle -6-2x = 4\sqrt{x+6}$?
  • Aug 5th 2011, 09:10 PM
  BERMES39
  Re: solving equations w/ radicals and exponents problem, stuck
  Try this,
  • Aug 5th 2011, 10:04 PM
  SammyS
  Re: solving equations w/ radicals and exponents problem, stuck
  To avoid producing extraneous solutions I like the following method.
  
  $-6-2x = 4\sqrt{x+6}$ , but first, divide both sides by 2 & get rid of the terms on the left side.
  
  Starting with: $0 = x+2\sqrt{x+6}+3$
  
  now add & subtract 6 at strategic locations: $0 =x+6+2\sqrt{x+6}+3-6$
  
  $0 = (\sqrt{x+6})^2+2\sqrt{x+6}-3$
  
  Now factor: $0 = (\sqrt{x+6}+3)(\sqrt{x+6}-1)$
  
  Since $\sqrt{x+6}\ne-3$, the only solution can be $\sqrt{x+6}=1\,.$
  
  This gives x=-5 .
  • Aug 6th 2011, 05:58 AM
  mrmohamed
  Re: solving equations w/ radicals and exponents problem, stuck
  Quote:
  
  Originally Posted by mathmathmathmathmathmathm
  -6 - 2x = 4 times sqrt of x + 6
  
  this was a part of a larger problem but i see this is where it screws up, the answer is -5 as the solution set
  
  what ive been doin w/ problems before this one was divide that 4 in both places and it always fit nicely but not in this problem, iam really frustrated can someone gimmie a hint? Thanks a bunch
  
  hi all