Question

is square root ... see equation

Answer 1

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

Answer 2

hm
that 2nd one is supposed to be root(2x) + root(3)

Answer 3

are you supposed to solve this for x?

Answer 4

no, im just asking if the stuff under the square root sign is commutative

Answer 5

on the left hand side

Answer 6

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

Answer 7

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

Answer 8

true if x = 0. otherwise forget it

Answer 9

yeah the latest one, is that true?

Answer 10

heck no

Answer 11

no. Think of it like this. \[\sqrt{9+16} = \sqrt{25}\] but \[\sqrt{9}=3 and \sqrt{16} = 4\] but 3+4\[\neq 5\] which is what \[\sqrt{25}\] is

Answer 12

suppose x = 12 then
\[\sqrt{2\times 12+3}=\sqrt{25}=5\]

Answer 13

whereas the right hand side would be
\[\sqrt{2\times 12}+\sqrt{3}=\sqrt{24}+\sqrt{3}\]

Answer 14

which is sure has heck not 25

Answer 15

ok