Question

Find values of s: \[\frac{s+\sqrt{s^2+2}}{2}\leq1\]

Answer 1

the radical has no restrictions ...

Answer 2

wat do u mean? Group Master?

Answer 3

if we do this normally ..... as if theres another way

s+sqrt(s^2 +2) </ 2

Answer 4

s aint got restrictions under the radical since when you square it it goes +

Answer 5

sqrt(s^2 +2) </ 2-s

Answer 6

answer would be s>=1/2 amistre

Answer 7

sorry s<=1/2

Answer 8

s^2 +2 </ 4 +s^2 -4s

2 </ 4-4s

-2 </ -4s

2>/ 4s

1/2 >/ s

Answer 9

yep; thats what I get :) without a dbl chk

Answer 10

\[\sqrt{s^2+2}\leq2-s \implies s^2+2\leq(s-2)^2\ for\ positive\ s.\]

Answer 11

\(\frac{-7}{4}\le s \le \frac{1}{2}\). That's what I got.

Answer 12

For negative s, let x=-s and
\[\sqrt{x^2+2}\leq2+x\implies x^2+2\leq4+x^2+4x\implies-2\leq4x\implies x\geq-1/2\implies s\leq1/2\]

Answer 13

i got x < 1/2

Answer 14

u get s<=1/2 always.

Answer 15

oops so did everyone else!

Answer 16

\[(\sqrt{s ^{2}+2})^{2}\le (2-s)^{2}\]
\[s^{2}+2\le4-4s+s ^{2}\]
\[4s-2\le0\]
\[s \le1/2 \forall s \in \mathbb{R}\]

Answer 17

\(s\le\frac{1}{2}\) is not enough condition. For example s=0 gives that \(\frac{\sqrt{2}}{2}\le\frac{1}{2}\), which is not right. So, I think that \(s\le-\frac{7}{4}\).