1. ## 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

2. ## Re: solving equations w/ radicals and exponents problem, stuck

heres the orginal problem

sqrt of 4 - x - sqrt of x + 6 = 2

3. ## Re: solving equations w/ radicals and exponents problem, stuck

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:
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: $-6-2x = 2\sqrt{x+6}$

4. ## Re: solving equations w/ radicals and exponents problem, stuck

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

Try this,

6. ## 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 .

7. ## Re: solving equations w/ radicals and exponents problem, stuck

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