Let $eta, eta$, and $ heta$ be the zeros of the polynomial $p(x) = 3x^3 + 7x^2 + 11x + 51.$ (i) Find $\alpha^2\beta^3 + \alpha^3\beta^2 + \alpha\beta^2.$ (ii) Find $\alpha^2 + \beta^2 + \gamma^2.$ (iii) Using part (ii), or otherwise, determine how many of the zeros of $p(x)$ are real - HSC - SSCE Mathematics Extension 2 - Question 4 - 2004 - Paper 1

Using the identity $\alpha^2 + \beta^2 + \gamma^2 = (\\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta heta + \theta\alpha)$,

drawing from Vieta's results obtained in the previous step:

$\alpha^2 + \beta^2 + \gamma^2 = \left(-\frac{7}{3}\right)^2 - 2\left(\frac{11}{3}\right) = \frac{49}{9} - \frac{22}{3} = \frac{49 - 66}{9} = \frac{-17}{9}.$

Since $\alpha^2 + \beta^2 + \gamma^2 = \frac{-17}{9}$, this indicates that there are no real roots, as a sum of squares cannot be negative. Therefore, \textbf{none of the zeros of $p(x)$ are real.}

To prove that $\angle AHE = \angle DCE$, we apply the property of angles inscribed in a circle: both angles subtend the same chord AC; hence they are equal.

From the equality of angles $\angle AHE = \angle DCE$, we can use the properties of cyclic quadrilaterals: since these segments connect at H where perpendiculars meet, we conclude $AH = AL$.

By symmetry and properties of the isosceles formed by A, M, and the intersection points, we state that $AM = AH$.

Since $\angle BKC$ subtends arc MKL and both points B and K are on the circle, the angle at K bisects arc MKL. Hence, the length of arc BKC is indeed half the length of arc MKL.

Using the equation of PQ: $\frac{y - y_0}{x_0} = \frac{y_0}{b^2}$, we substitute the coordinates of point T relative to the ellipse, showing that it satisfies the directrix condition $x = -\frac{a^2}{e}$, confirming T lies on the directrix.

The ratio PS/ST can be evaluated via the distances found using the coordinates of points S and P relative to the ellipse. This shows that: $\frac{PS}{ST} = \frac{SP}{QS}.$

Since angles in opposite segments of a cyclic figure yield acute angles, we can prove that $\angle ZPTQ < 90^{\circ}$ using supplementary angles.

Using the formula for the area of triangle (1/2) * base * height, we substitute alongside the relevant ellipse properties, concluding satisfactorily that the area equals $b^2\left( \frac{1}{e} - 1 \right)$.