Question

√x+4+2=x

Answer 1

\[x-\sqrt{x}-6=0\]
solve it as \[consider x= (\sqrt{x})^{2}\]
solve it as quadratic equation

Answer 2

\[(\sqrt{x})^{2}-\sqrt{x}-6=0\]

Answer 3

\[\sqrt{x}=3 \] (only one positive root is available)
then x=9

Answer 4

how did you get √x-6

Answer 5

4+2=6

Answer 6

the original problem was
\[\sqrt{x}+4+2=x\]
\[\sqrt{x}+6=x\]
\[x-\sqrt{x}-6=0\]
see solution up the page

Answer 7

are you OK now....?

Answer 8

Yes lol thanks i was was confused

Answer 9

Great!
it was a tricky one!

Answer 10

inik, why is the solution x=4 not considered. If the - root of 4 a -2 is considered then
-2+4+2=4
4=4
just curious...

Answer 11

I got sq. rt (x) = (1+/-5)/2
as a solution
x=3 is only positive root
x=-2 is root too, but
sq/rt (x) can not be = -2 for real numbers and I assumed that for this level they dont use imaginary numbers...
am i wrong?

Answer 12

\[\sqrt{4}+4+2=4\]

Answer 13

You should test in the original problem. It is not taking a root of negative number, this is not a complex number it is simplly a 4.

Answer 14

yes, but the original equation has:
\[\sqrt{-2}...\]
i'm not sure he can deal with that one

Answer 15

Where is there a \[\sqrt{-2}??\]

Answer 16

i see...

Answer 17

Then is 4 a good answer?

Answer 18

let's see:
\[\sqrt{4}+4+2=2+4+2=8\]
right?
8 is not 4
now a solution

Answer 19

i mean not a solution.... :)

Answer 20

I think when you squared it you took away the sign. I didn't say 8 was a solution I am talking asbout a simple positive 4

Answer 21

i used x=4....in original eq

Answer 22

No the square root of 4 is a + or - 2 Why are you excluding the -2 as the square root of X? Nothing imaginary about it.

Answer 23

I understand why you excluded the positive root.

Answer 24

Is that why you excluded the 4 because the positive root gives a false answer?

Answer 25

yes

Answer 26

I guess that makes sense.

Answer 27

when you look at 9 as a solution
\[\sqrt{9}+4+2=9\] the the negative root was ignored too. It is just worrisome to me to ignore one root. I guess the context of the problem will guide you to determine the correct one.