Let $eta$ and $eta$ be the zeros of the polynomial $p(x) = 3x^3 + 7x^2 + 11x + 51$ - HSC - SSCE Mathematics Extension 2 - Question 4 - 2004 - Paper 1

Using the relationship between the roots and coefficients from Vieta's formulas, we know:

$\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha \beta + \beta \gamma + \gamma \alpha)$

Substituting these values leads to the calculated result.

To determine the number of real zeros, analyze the calculated value from part (ii). Utilizing the discriminant $D = b^2 - 4ac$, we can establish whether the function intersects the x-axis and consequently deduce how many zeros are real.

From the properties of cyclic quadrilaterals, particularly that the angles subtended at the circle's circumference related to the same arc are equal, we can articulate the proof that $\angle AHE = \angle ZDC$ $.

With the established equality of angles from part (i), we invoke the isosceles triangle theorem which implies that the sides opposite equal angles are also equal, thus providing $AH = AL$.

For triangle $AMH$, we can conclude that $AM = AH$ following the same logic of equal angles within that respective triangle, using properties of cyclic triangles.

We know from circle properties that the measure of an arc is proportional to the angle subtended at the center. Thus, since $\angle BOC = \angle MOK$ (which is equal to half $\angle BKC$ from inscribed angle theorem), it follows that: $\text{length}(BKC) = \frac{1}{2} \text{length}(MKL).$

To determine whether the point $T$ lies on the directrix, we can substitute the coordinates derived from the equation of $PQ$ into that of the directrix equation $X = \frac{a^2}{x_0}$ and simplify accordingly, demonstrating that $T$ satisfies both conditions.

To calculate the ratio $PS:ST$, apply the distance formula between the points $P$ and $S$ as well as between $S$ and $T$. Express it in terms of the known coordinates, leading to: $\frac{PS}{ST} = \frac{\text{distance from } P \text{ to } S}{\text{distance from } S \text{ to } T}$.

To demonstrate that angle $\angle ZPTQ$ is acute, we analyze the slopes of the tangents at points $P$ and $Q$. The calculations will show that the product of their slopes must be less than -1 thereby proving that the angle formed is indeed less than 90 degrees.

Utilizing the formula for the area of triangle formed by the points $P$, $Q$, and $T$, we substitute the respective coordinates into the formula: