SSCE HSC Mathematics Extension 2 Polynomial equations: a) Prove that $ \sqrt{23} $ is

Question

a) Prove that $ \sqrt{23} $ is irrational.

Assume the contrary that $ \sqrt{23} $ is rational. Then it can be expressed as $ \sqrt{23} = \frac{p}{q} $ where $ p $ and $ q $ are coprime integers, and $ q \neq 0 $.

Squaring both sides gives:
$$ 23 = \frac{p^2}{q^2} $$
$$ 23q^2 = p^2 $$

This implies that $ p^2 $ is divisible by 23, so $ p $ must also be divisible by 23.

Let $ p = 23k $ for some integer $ k $. Substituting this into our equation, we have:
$$ 23q^2 = (23k)^2 = 529k^2 $$
$$ q^2 = 23k^2 $$

Thus, $ q^2 $ is also divisible by 23 and consequently, $ q $ is divisible by 23. Therefore, both $ p $ and $ q $ share a common factor of 23, contradicting our assumption that they are coprime. Hence, $ \sqrt{23} $ is irrational.

b) Prove that for all real numbers $ x $ and $ y $, where $ x^2 + y^2 \neq 0 $:
$$ \frac{(x + y)^2}{x^2 + y^2} \leq 2. $$

Starting with the left side:
$$ \frac{(x + y)^2}{x^2 + y^2} = \frac{x^2 + 2xy + y^2}{x^2 + y^2} $$

Let $ k = x^2 + y^2 $, we can rewrite the equation as:
$$ = 1 + \frac{2xy}{k}. $$

By the AM-GM inequality, we have:
$$ \frac{x^2 + y^2}{2} \geq \sqrt{x^2y^2} \rightarrow x^2 + y^2 \geq 2xy. $$

Thus:
$$ \frac{2xy}{k} \leq 1. $$

This yields:
$$ 1 + \frac{2xy}{k} \leq 2. $$

c) An object with mass $ m $ kilograms slides down a smooth inclined plane with velocity $ y(t) $, where $ t $ is the time in seconds after the object started sliding down the plane. The inclined plane makes an angle $ \theta $ with the horizontal, as shown in the diagram. The normal reaction force is $ R $ and has magnitude $ g $. No other forces act on the object.

(i) Show that the resultant force on the object is $ \mathbf{F} = - (mg \sin \theta) \mathbf{j} $.

Resolving the weights:
- Perpendicular to the slope gives:
$$ R = mg \cos \theta. $$
- Parallel to the slope gives:
$$ mg \sin \theta = mg \sin \theta. $$
$$ \text{Resultant Force} = \mathbf{F} = - (mg \sin \theta) \mathbf{j}. $$

(ii) Given that the object is initially at rest, find its velocity $ y(t) $ in terms of $ g, \theta, t $, and $ L $.

Using the relation for acceleration:
$$ mg - mg \sin \theta = ma \rightarrow a = g \sin \theta. $$

Integrating:
$$ v = g \sin \theta t. $$

d) Find the cube roots of $ 2 - 2i $. Give your answer in exponential form.

Let $ z = 2 - 2i = r e^{i\theta}, \text{ where } r = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2} \text{ and } \theta = \tan^{-1}\left(\frac{-2}{2}\right) = -\frac{\pi}{4}. $

Thus:
$$ z = 2\sqrt{2} e^{-i\frac{\pi}{4}}. $$

The cube roots are:
$$ \sqrt[3]{2\sqrt{2}} e^{i\left(-\frac{\pi}{4} + \frac{2k\pi}{3}\right)}, k = 0, 1, 2. $$

e) The complex number $ 2 + i $ is a zero of the polynomial $ P(z) = z^4 - 3z^3 + cz^2 + dz - 30 $ where $ c $ and $ d $ are real numbers.

(i) Explain why $ 2 - i $ is also a zero of the polynomial $ P(z) $.

Since $ P(z) $ has real coefficients, zeros occur in complex conjugate pairs.
Thus, if $ 2 + i $ is a zero, then $ 2 - i $ is also a zero.

(ii) Find the remaining zeros of the polynomial $ P(z) $.

a) Prove that \( \sqrt{23} \) is irrational - HSC - SSCE Mathematics Extension 2 - Question 12 - 2023 - Paper 1

Worked Solution & Example Answer:a) Prove that \( \sqrt{23} \) is irrational - HSC - SSCE Mathematics Extension 2 - Question 12 - 2023 - Paper 1

Prove that \( \sqrt{23} \) is irrational.

Prove that for all real numbers \( x \) and \( y \), where \( x^2 + y^2 \neq 0 \): \( \frac{(x + y)^2}{x^2 + y^2} \leq 2. \)

Show that the resultant force on the object is \( \mathbf{F} = - (mg \sin \theta) \mathbf{j} \).

Given that the object is initially at rest, find its velocity \( y(t) \) in terms of \( g, \theta, t, \) and \( L \).

Find the cube roots of \( 2 - 2i \). Give your answer in exponential form.

Explain why \( 2 - i \) is also a zero of the polynomial \( P(z) \).

Find the remaining zeros of the polynomial \( P(z) \).

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