The equation $4k^3 - 27k + k = 0$ has a double root - HSC - SSCE Mathematics Extension 2 - Question 5 - 2002 - Paper 1

Given the polynomial $P(x) = x^3 - 5x^2 + 5$, we want to find a polynomial whose roots are $\alpha - 1, \beta - 1$, and $\gamma - 1$. To accomplish this, we can perform a substitution $x = y + 1$, yielding: $P(y+1) = (y+1)^3 - 5(y+1)^2 + 5$ Expanding this gives: $= y^3 + 3y^2 + 3y + 1 - 5(y^2 + 2y + 1) + 5 = y^3 - 2y + 1$ Thus, the polynomial with the desired roots is: $Q(y) = y^3 - 2y + 1 = 0$

To find the polynomial with roots $\alpha^2, \beta^2$, and $\gamma^2$, we can similarly transform the original polynomial $P(x)$ using the substitution $x = \sqrt{y}$.We know: $$P(\sqrt{y}) = (\sqrt{y})^3 - 5(\sqrt{y})^2 + 5 = 0$$ Lettingx = \sqrt{y}gives: $$x^3 - 5x^2 + 5 = 0$$ Squaring the roots gives us: $$y^3 - 5y + 5 = 0$$ Thus, the desired polynomial whose roots are\alpha^2, \beta^2,$and$\gamma^2$ is: $R(y) = y^3 - 5y + 5 = 0$

To find the value of $\alpha^3 + \beta^3 + \gamma^3$, we can use the identity: $\alpha^3 + \beta^3 + \gamma^3 = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha) + 3\alpha\beta\gamma$ From the original polynomial $x^3 - 5x^2 + 5 = 0$, we determine:

• $\alpha + \beta + \gamma = 5$
• $\alpha\beta + \beta\gamma + \gamma\alpha = 0$
• $\alpha\beta\gamma = -5$ Substituting these values gives: $\alpha^3 + \beta^3 + \gamma^3 = 5(\alpha^2 + \beta^2 + \gamma^2) + 3(-5)$ We also know that: $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = 25 - 0 = 25$ Thus: $\alpha^3 + \beta^3 + \gamma^3 = 5(25) - 15 = 125 - 15 = 110$

To show that the equation of the tangent to the ellipse at point $P(\frac{x_i}{y_i})$ is given by: $\frac{x_i^2}{a^2} + \frac{y_i^2}{b^2} = 1,$ it is essential to derive the tangent line equation from the standard form of the ellipse. Start with the blend of differential calculus and implicit differentiation: $\frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$ Using known points $(x_i, y_i)$ where they satisfy the original ellipse equation leads to the relation for the tangent as: $y - y_i = \frac{dy}{dx}(x - x_i)$ If you rearrange this equation to express the tangent line mathematically has just one such equation providing it's a point of tangency.

To derive the chord of contact from point $T(\frac{x_0}{y_0})$, we utilize the property of tangents from external points. The equation of the chord of contact from the point $T$ to the ellipse is given by: $\frac{x_0}{a^2} + \frac{y_0}{b^2} = 1.$ Thus, we confirm the relationship that obtains the chord subtractively assessing contact with the ellipse ensures direct connection as required.