Solve the inequation: 4x^2+ 8x + 3 > 0
We'll use another method to determine the interval of x values where the given expression is positive.
According to the rule, between the roots of the expression 4x^2+ 8x + 3, the expression has the opposite sign of the coefficient of x^2 and outside the roots, the expression has the same sign as the coefficient of x^2.
We'll calculate the roots:
4x^2+ 8x + 3 = 0
We'll apply the quadratic formula:
x1 = [-8 + sqrt(64-48)]/8
x1 = (-8 + 4)/8
x1 = -1/2
x2 = (-8-4)/8
x2 = -3/2
So, the expression is positive over the intervals:
First we find the roots of the equation 4x^2+8x+3 = 0. And then we decide the inequality 4x^2+8x+3 > 0.
4x^2+8x+3 = 0 .
We factorise the left:
4x^2+6x+2x+3 = 0
2x(x+3)+1(2x+3) = 0
(x+3)(2x+1) = 0
(x+3) (2x+1) = 0
The roots are : x = -3 and x = -1/2
So x = -3 or x =-1/2 are the values which make f(x) = 4x^2+8x+3 = (x+3)(2x+1/2) = 2(x+3)(x+1/2) = 0.
If x < -3 , then 2(x+3)(x+1/2) is 2(-ve)(-ve) is positive. So f(x) > 0.
If x > -1/2, then both factors x+3 and x+1/2 are postive. So their product is positive. So f(x) positive or f(x) > 0.
Therefore x should not belong to the inter val (-3 , -1/2) and should be outside this interval for f(x) = 4x^2+8x+3 to be greater than zero.
We are given the following inequation to solve 4x^2+ 8x + 3 > 0
Now, 4x^2+ 8x + 3 > 0
=> 4x^2 + 6x + 2x +3 >0
=> 2x ( 2x +3) + 1 (2x+3) >0
Now as (2x+1)(2x+3) is greater than 0. Either (2x+1)and (2x+3) can be greater than 0 or they can both be less than 0.
If they are greater than 0,
=> (2x+1) > 0 and (2x+3) >0
=> x > -1/2 and x >-3/2
=> x has to be greater than -1/2
If both (2x+1)and (2x+3) are less than 0
=> (2x+1) <0 and (2x+3) <0
=> x < -1/2 and x <-3/2
=> x has to be less than -3/2
Therefore x can be either greater than -1/2 or less than -3/2