# solving equations w/ radicals and exponents problem, stuck

• Aug 5th 2011, 12:03 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, 12:04 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, 12:07 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:

$\displaystyle (-6-2x)^2 = (4\sqrt{x+6})^2 \ \Leftrightarrow \ 4x^2+24x+36 = 16(x+6)$

Solve for x and check for extraneous solutions using the original equation. Note that due to the domain $\displaystyle x \geq -6$

EDIT: Your equation in the OP is fine but there is an additional domain restriction $\displaystyle -6 \leq x \leq 4$

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

$\displaystyle \sqrt{4 - x} - \sqrt{x + 6} = 2$

$\displaystyle \sqrt{4-x} = 2 + \sqrt{x+6}$

$\displaystyle 4-x = 4 + 4\sqrt{x+6} + x+6 = 10 + x + 4\sqrt{x+6}$

If you tidy that up a bit: $\displaystyle -6-2x = 2\sqrt{x+6}$
• Aug 5th 2011, 06:45 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 $\displaystyle \displaystyle -6-2x = 4\sqrt{x} + 6$ or $\displaystyle \displaystyle -6-2x = 4\sqrt{x+6}$?
• Aug 5th 2011, 08:10 PM
BERMES39
Re: solving equations w/ radicals and exponents problem, stuck
Try this,
• Aug 5th 2011, 09:04 PM
SammyS
Re: solving equations w/ radicals and exponents problem, stuck
To avoid producing extraneous solutions I like the following method.

$\displaystyle -6-2x = 4\sqrt{x+6}$ , but first, divide both sides by 2 & get rid of the terms on the left side.

Starting with: $\displaystyle 0 = x+2\sqrt{x+6}+3$

now add & subtract 6 at strategic locations: $\displaystyle 0 =x+6+2\sqrt{x+6}+3-6$

$\displaystyle 0 = (\sqrt{x+6})^2+2\sqrt{x+6}-3$

Now factor: $\displaystyle 0 = (\sqrt{x+6}+3)(\sqrt{x+6}-1)$

Since $\displaystyle \sqrt{x+6}\ne-3$, the only solution can be $\displaystyle \sqrt{x+6}=1\,.$

This gives x=-5 .
• Aug 6th 2011, 04:58 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