Question

Solve.

Square root of x+6=2-square root of x+2

I know you square both sides and get

x+6=(2-square root x+2)(2-square root x+2)

Help please! I'm stuck the answer is x= -2
But I don't know all the steps

Answer 1

is it
\[x+6=2-\sqrt{x+2}\]?

Answer 2

if so first step is subtract 2 to get
\[x+4=\sqrt{x+2}\] then square

Answer 3

oh okay & then you...
square both sides?

Answer 4

wait no
the x+6 is a square root

Answer 5

so you can't subtract the 2 right? It'd be x+6=4-2squareroot x+2 - 2squareroot x+2+x+2

Answer 6

well then you have to square both sides twice

Answer 7

yes after doing that, I got what I just sent you is that right?

Answer 8

\[x+6=4-4\sqrt{x+2}+x+2\]

Answer 9

that is step one yes?

Answer 10

Yes and then I would add the 2 to the 4 so then it would equal

x+6=6+x-4squareroot x+2 right?

Answer 11

now we get
\[0=-4\sqrt{x+2}\] making \[\sqrt{x+2}=0\] or \[x+2=0\]
\[x=-2\]

Answer 12

you have an extra 2 hanging out at the end i think. 4+2 = 6

Answer 13

Oh okay I get it! :D

Answer 14

you wrote

Yes and then I would add the 2 to the 4 so then it would equal

x+6=6+x-4squareroot x+2 right?

but i think you just get
x+6=6+x-4 root(x+2)

Answer 15

so answer is 0 and check is easy

Answer 16

I have more if you have time? I have a test tomorrow :(

Answer 17

yes I see where I messed up thank you

Answer 18

\[\sqrt{2x+2}-\sqrt{x-3}=2\]

Answer 19

I know you take x square root x-3 to the other side first

Answer 20

so then it would be\[\sqrt{2x+2}^{2}=(2+\sqrt{x-3)^{2}}\]