sqr(3-x)-sqr(1+x) >1/2.......................find x

We'll impose constraints of existence of square
roots:

3 - x >= 0

x
=< 3

1 + x >=
0

x >= -1

The range of
admissible values for x is [-1 ; 3].

Now, we'll solving the
inequality by raising to square both sides:

3 - x + 1 + x -
2sqr(3-x)*(1+x) > 1/4

We'll eliminate x and we'll
combine like terms:

4 - 2sqr(3-x)*(1+x) >
1/4

We'll subtract 4 both
sides:

- 2sqr(3-x)*(1+x) > 1/4 -
4

- 2sqr(3-x)*(1+x) >
-15/4

We'll multiply by
-1:

2sqr(3-x)*(1+x) <
15/4

We'll divide by
2:

sqr(3-x)*(1+x) <
15/8

We'll raise to square
again:

(3-x)*(1+x) <
225/64

We'll subtract
225/64:

(3-x)*(1+x) - 225/64 <
0

We'll remove the brackets:

3
+ 2x - x^2 - 225/64 < 0

64*3 + 64*2x - 64x^2 - 225
< 0

- 64x^2 + 128x - 33 <
0

64x^2 - 128x + 33 >
0

The expresison above is strictly positive if discriminant
of the quadratic is negative:

delta = 16384 -
8448

delta = 7936

Since delta
>0, the quadratic has 2 real roots.

x1 =
(128+16sqrt31)/2*64

x1 =
16(8+sqrt31)/2*2*2*16

x1 =
(8+sqrt31)/8

x2 =
(8-sqrt31)/8

The expression is positive over
the ranges [-1;(8-sqrt31)/8) U ((8+sqrt31)/8 ;
3].