the equation can be written as
4x^2 +4x -3 > 0

(2x-1)(2x+3)> 0

Either both 2x-1 and 2x+3 are +ve or both are -ve.

1. both +ve
2x-1 > 0
x> 1/2....... equ(a)

2x+3 > 0
x> -3/2....... equ(b)

a and b gives x > 1/2

2. Bothe -ve

2x-1 < 0
x< 1/2 ........ equ(c)

2x+3 < 0
x< -3/2 .......... equ(d)

c and d gives x < -3/2


SO, x>=1/2 and x=< -3/2

mariofelixpasku wrote:how do you get the values of x if you got sthg like this:

4x(x+1) ≥ 3

answers: x>=1/2 and x=< -3/2

Step 1: Rephrase the inequality in terms of 0 and factor the quadratic.
4x(x+1) ≥ 3
4x² + 4x - 3 ≥ 0
(2x + 3)(2x - 1) ≥ 0

Step 2: Test one value to the left and one value to the right of each CRITICAL POINT.
The critical points are the values of x where the left-hand side is equal to 0.
Here, the critical points are x = -3/2 and x = 1/2.
These two points are valid solutions because, in each case, (2x + 3)(2x - 1) = 0.
To determine the range(s) where (2x + 3)(2x - 1) > 0, test one value to each side of x=-3/2 and x=1/2.

Plug x=-2 -- a value to the LEFT of -3/2 -- into (2x + 3)(2x - 1) ≥ 0 :
( 2(-2) + 3 ) ( 2(-2) - 1) ≥ 0
(-1)(-5) ≥ 0
5 ≥ 0.
This works.
x < -3/2 is a viable range.

Plug x=0 -- a value BETWEEN -3/2 and 1/2 -- into (2x + 3)(2x - 1) ≥ 0 :
(2*0 + 3)(2*0 - 1) ≥ 0
(3)(-1) ≥ 0
-3 ≥ 0.
Doesn't work.
-3/2 < x < 1/2 is not a viable range.

Plug x=1 -- a value to the RIGHT of 1/2 -- into (2x + 3)(2x - 1) ≥ 0 :
(2*1 + 3)(2*1 - 1) ≥ 0
(5)(1) ≥ 0
5 ≥ 0.
This works.
x > 1/2 is a viable range.

Thus, the viable ranges are x≤-3/2 and x≥1/2.