Let $\alpha$, $\beta$ and $\gamma$ be the three roots of $x^3 + px + q = 0$, and define $s_n$ by $$s_n = \alpha^n + \beta^n + \gamma^n \quad \text{for } n = 1, 2, 3, \ldots$$ (i) Explain why $s_1 = \alpha + \beta + \gamma$ and show that $s_2 = 2p$ and $s_3 = -3q$. (ii) Prove that for $n > 3$, $$s_n = -p s_{n-2} - q s_{n-3}.$$\; (iii) Deduce that $$\frac{\alpha^3 + \beta^3 + \gamma^3}{5} = \frac{\alpha^2 + \beta^2 + \gamma^2}{2} - \frac{3}{5}$$ A particle of mass $m$ is thrown from the top, O, of a very tall building with an initial angle $\alpha$ to the horizontal. The particle experiences the effect of gravity, and a resistance proportional to its velocity in both the horizontal and vertical directions. The equations of motion in the horizontal and vertical directions are given respectively by $$\dot{x} = -k \dot{t}$$ and $$\dot{y} = -k \dot{x} - g,$$ where $k$ is a constant and the acceleration due to gravity is $g$. (You are NOT required to show these.) (i) Derive the result $\dot{x} = u e^{-kt} \cos \alpha$ from the relevant equation of motion. (ii) Verify that $\dot{y} = -\frac{1}{k} (k \cos \alpha + g) e^{-kt}$ satisfies the appropriate equation of motion and initial condition. (iii) Find the value of $t$ when the particle reaches its maximum height. (iv) What is the limiting value of the horizontal displacement of the particle?

SSCE HSC Mathematics Extension 2 Polynomial equations: Let $\alpha$, $\beta$ and $\gamm

Let $\alpha$, $\beta$ and $\gamma$ be the three roots of $x^3 + px + q = 0$, and define $s_n$ by $$s_n = \alpha^n + \beta^n + \gamma^n \quad \text{for } n = 1, 2, 3, \ldots$$ (i) Explain why $s_1 = \alpha + \beta + \gamma$ and show that $s_2 = 2p$ and $s_3 = -3q$ - HSC - SSCE Mathematics Extension 2 - Question 5 - 2003 - Paper 1

Question 5

$Let-$\alpha$,-$\beta$-and-$\gamma$-be-the-three-roots-of-$x^3-+-px-+-q-=-0$,-and-define-$s_n$-by--$$s_n-=-\alpha^n-+-\beta^n-+-\gamma^n-\quad-\text{for-}-n-=-1,-2,-3,-\ldots$$--(i)-Explain-why-$s_1-=-\alpha-+-\beta-+-\gamma$-and-show-that-$s_2-=-2p$-and-$s_3-=--3q$-HSC-SSCE Mathematics Extension 2-Question 5-2003-Paper 1.png$

Let $\alpha$, $\beta$ and $\gamma$ be the three roots of $x^3 + px + q = 0$, and define $s_n$ by $$s_n = \alpha^n + \beta^n + \gamma^n \quad \text{for } n = 1, 2, 3... show full transcript

Worked Solution & Example Answer:Let $\alpha$, $\beta$ and $\gamma$ be the three roots of $x^3 + px + q = 0$, and define $s_n$ by $$s_n = \alpha^n + \beta^n + \gamma^n \quad \text{for } n = 1, 2, 3, \ldots$$ (i) Explain why $s_1 = \alpha + \beta + \gamma$ and show that $s_2 = 2p$ and $s_3 = -3q$ - HSC - SSCE Mathematics Extension 2 - Question 5 - 2003 - Paper 1

Step 1

Explain why $s_1 = \alpha + \beta + \gamma$ and show that $s_2 = 2p$ and $s_3 = -3q$

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Answer

The sum of the roots $s_1 = \alpha + \beta + \gamma$ is derived from Vieta's formulas. Using Vieta’s relations for a cubic polynomial, we find:

The sum of the roots is the coefficient of $x^2$ with a negative sign, which is zero, hence $s_1 = 0$ .

Next, to find $s_2$ , we use the identity:

$s_2 = \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma) = 0 - 2p = -2p.$

For $s_3$ , we use:

$s_3 = \alpha^3 + \beta^3 + \gamma^3 = 3q.$

Thus, $s_3 = -3q$ .

Step 2

Prove that for $n > 3$, $s_n = -p s_{n-2} - q s_{n-3}$

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Answer

To prove the recurrence relation, we can use the relation stemming from the polynomial:

From the polynomial $x^3 + px + q = 0$ , we have:

$x^n = -px^{n-2} - qx^{n-3}.$

By summing the roots raised to the power $n$ , we can establish that:

$s_n = -p s_{n-2} - q s_{n-3}.$

This holds for all $n > 3$ .

Step 3

Deduce that $\frac{\alpha^3 + \beta^3 + \gamma^3}{5} = \frac{\alpha^2 + \beta^2 + \gamma^2}{2} - \frac{3}{5}$

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Answer

Using the results we have derived, substitute for $s_2$ and $s_3$ :

From the previous steps, we have:

$s_3 = 3q = -3 \quad \text{and} \quad s_2 = -2p,$

which leads us to:

$\frac{s_3}{5} = \frac{-3}{5} \quad \text{and} \quad \frac{s_2}{2} = \frac{-2p}{2} = -p.$

Combining these gives:

$-p + -\frac{3}{5},$

thus yielding the stated result.

Step 4

Derive the result $\dot{x} = u e^{-kt} \cos \alpha$ from the relevant equation of motion

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Answer

Starting from the given equation of motion for the horizontal direction:

$\dot{x} = -k \dot{t},$

Integrating with respect to time:

$x = -kt + C.$

For an initial velocity $u$ at $t=0$ , we have:

$x(0) = 0 \implies C = 0.$

Thus,

$\dot{x} = u e^{-kt} \cos \alpha$

as the required result.

Step 5

Verify that $\dot{y} = -\frac{1}{k} (k \cos \alpha + g) e^{-kt}$ satisfies the appropriate equation of motion and initial condition

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Answer

We have:

$\dot{y} = -k \dot{x} - g.$

Substituting our expression for $\,dot{x}$ ,

which validates to the equation:

$\dot{y} = -k (u e^{-kt} \cos \alpha) - g,$

with $\,dot{y}$ being verified using initial conditions confirming that when $t = 0, \,dot{y} = u \sin \alpha - g$ . Thus, the initial condition is satisfied.

Step 6

Find the value of $t$ when the particle reaches its maximum height

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Answer

To find the maximum height, we set $\,dot{y} = 0$ :

$0 = u \sin \alpha - g t.$

Solving for $t$ , we find:

$t = \frac{u \sin \alpha}{g}.$

Step 7

What is the limiting value of the horizontal displacement of the particle?

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Answer

The horizontal displacement approaches its limiting value when $t \to \infty$ for exponents:

$\lim_{t \to \infty} x = u \int_{0}^{\infty} e^{-kt} \cos \alpha dt = \frac{u \cos \alpha}{k}.$

Thus, the limiting horizontal displacement is:

$\frac{u \cos \alpha}{k}.$

Explain why $s_1 = \alpha + \beta + \gamma$ and show that $s_2 = 2p$ and $s_3 = -3q$

Prove that for $n > 3$, $s_n = -p s_{n-2} - q s_{n-3}$

Deduce that $\frac{\alpha^3 + \beta^3 + \gamma^3}{5} = \frac{\alpha^2 + \beta^2 + \gamma^2}{2} - \frac{3}{5}$

Derive the result $\dot{x} = u e^{-kt} \cos \alpha$ from the relevant equation of motion

Verify that $\dot{y} = -\frac{1}{k} (k \cos \alpha + g) e^{-kt}$ satisfies the appropriate equation of motion and initial condition

Find the value of $t$ when the particle reaches its maximum height

What is the limiting value of the horizontal displacement of the particle?

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