It is given that $a$ and $b$ are positive constants such that $3b > a^2$. Explain why $y = x^2 + 2ax + 2a^2 - 3b$ has two $x$-intercepts for all real values of $x$. The curve $y^2 + y - 3x = 9$ and the line $2x = 5 - y$ intersect at the points $P$ and $Q$. Find the $x$-coordinate of $P$ and $Q$.

1. Set up the discriminant for the quadratic equation when finding $x$-intercepts.

To find the $x$-intercepts, we set $y = 0$:
$x^2 + 2ax + 2a^2 - 3b = 0$
The discriminant is given by $\Delta = B^2 - 4AC$.

2. Calculate and simplify the discriminant.

$\Delta = (2a)^2 - 4(1)(2a^2 - 3b)$
$\Delta = 4a^2 - 8a^2 + 12b$
$\Delta = 12b - 4a^2$
$\Delta = 4(3b - a^2)$

3. Use the given condition to determine the sign of the discriminant and conclude.

We are given that $3b > a^2$, which means $3b - a^2 > 0$.
Therefore, $\Delta = 4(3b - a^2) > 0$.
Since the discriminant is strictly positive, the quadratic equation has two distinct real roots. Thus, the curve has two $x$-intercepts.

1. Express $y$ in terms of $x$ from the linear equation

From $2x = 5 - y$, we have:
$y = 5 - 2x$

2. Substitute the expression for $y$ into the non-linear equation

Substitute $y = 5 - 2x$ into $y^2 + y - 3x = 9$:
$(5 - 2x)^2 + (5 - 2x) - 3x = 9$

3. Expand and simplify to form a quadratic equation in $x$

$(25 - 20x + 4x^2) + 5 - 2x - 3x = 9$
$4x^2 - 25x + 30 = 9$
$4x^2 - 25x + 21 = 0$

4. Solve the quadratic equation for $x$

$(4x - 21)(x - 1) = 0$
$x = \frac{21}{4} = 5.25$ or $x = 1$