9x2-6xy+4y2>0

To solve the inequality \( 9x^2 - 6xy + 4y^2 > 0 \), we can treat it as a quadratic in terms of \( x \):

\[ 9x^2 - 6xy + 4y^2 > 0 \]

This can be written in standard quadratic form: \[ ax^2 + bx + c > 0 \] where:

• \( a = 9 \)
• \( b = -6y \)
• \( c = 4y^2 \)

Next, we will consider the discriminant of this quadratic, which determines the nature of the roots. The discriminant \( D \) is given by: \[ D = b^2 - 4ac \] Substituting our values: \[ D = (-6y)^2 - 4(9)(4y^2) = 36y^2 - 144y^2 = -108y^2 \]

The discriminant \( D = -108y^2 \) is negative for \( y \neq 0 \) (since \( y^2 \) is always non-negative and equals zero when \( y = 0 \)).

Since the discriminant is negative, this means the quadratic \( 9x^2 - 6xy + 4y^2 \) has no real roots and opens upwards (as \( a > 0 \)). Thus, it will always be positive for any \( x \) when \( y \neq 0 \).

Now let's examine the case where \( y = 0 \): \[ 9x^2 > 0 \quad \text{for } x \neq 0 \] This inequality holds true for all \( x \) except when \( x = 0 \).

In summary:

• For \( y \neq 0 \), \( 9x^2 - 6xy + 4y^2 > 0 \) holds for all \( x \).
• For \( y = 0 \), \( 9x^2 > 0 \) holds for all \( x \) except \( x = 0 \).

Thus, the solution to the inequality \( 9x^2 - 6xy + 4y^2 > 0 \) is:

• For \( y \neq 0 \): all \( x \in \mathbb{R} \)
• For \( y = 0 \): \( x \in \mathbb{R} \setminus \{0\} \)

This can be summarized as:

• The inequality holds for all \( (x, y) \) pairs except when \( y = 0 \) and \( x = 0 \).