# Solve the inequation:  4x^2+ 8x + 3 > 0

giorgiana1976 | Student

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

﻿﻿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:

(-inf., -3/2)U(-1/2,+inf.)

neela | Student

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.

william1941 | Student

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

=> (2x+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