The problem is to analyze the inequality $4x^2 + 8xy + 3y^2 > 0$. We want to determine the conditions on $x$ and $y$ that make this inequality true.

1. Problem Description

The problem is to analyze the inequality $4x^2 + 8xy + 3y^2 > 0$. We want to determine the conditions on $x$ and $y$ that make this inequality true.

2. Solution Steps

We can factor the quadratic expression:

$4x^2 + 8xy + 3y^2 = (2x + y)(2x + 3y)$

So, we have the inequality $(2x + y)(2x + 3y) > 0$.

This inequality holds if both factors are positive or both factors are negative.

Case 1: Both factors are positive.

$2x + y > 0$ and $2x + 3y > 0$

$y > -2x$ and $3y > -2x$, so $y > -\frac{2}{3}x$

In this case, since $-2 < -\frac{2}{3}$ is false, we need both inequalities to hold, so $y > -\frac{2}{3}x$.

Case 2: Both factors are negative.

$2x + y < 0$ and $2x + 3y < 0$

$y < -2x$ and $3y < -2x$, so $y < -\frac{2}{3}x$

In this case, since $-2 < -\frac{2}{3}$, we need both inequalities to hold, so $y < -2x$.

However, if $y=0$, the inequality becomes $4x^2 > 0$, which is true for $x \ne 0$. Thus, $x$ can be non-zero when $y=0$.

Combining these cases, we have $y > -\frac{2}{3}x$ or $y < -2x$.

Notice that if $x = 0$, then $3y^2 > 0$ implies $y \neq 0$.

If $y = 0$, then $4x^2 > 0$ implies $x \neq 0$.

Also, if $y = -x$, we have $4x^2 - 8x^2 + 3x^2 = -x^2 > 0$, which is never true.

If $y = -2x$, we have $4x^2 - 16x^2 + 12x^2 = 0$, which is not greater than

0. If $y = -\frac{2}{3}x$, we have $4x^2 - \frac{16}{3}x^2 + \frac{4}{3}x^2 = \frac{12 - 16 + 4}{3}x^2 = 0$, which is not greater than

0. Therefore the inequality holds if $y > -\frac{2}{3}x$ or $y < -2x$. We must also exclude the lines $y = -\frac{2}{3}x$ and $y=-2x$. Also exclude the point $(0,0)$.

3. Final Answer

The inequality $4x^2 + 8xy + 3y^2 > 0$ holds if $y > -\frac{2}{3}x$ or $y < -2x$. We must exclude the lines $y = -\frac{2}{3}x$ and $y=-2x$. Also, we must exclude the point (0,0).