QUESTION 3
The roots of a quadratic equation are .
3.1 Calculate the value(s) of for which the roots will be real.
3.2 Give one value of for which the roots will be rational.
[3]

QUESTION 4
Simplify the following without the use of a calculator:

[13]

QUESTION 5
5.1 If and , determine with the aid of a diagram the value of .
5.2 If , write down the following in terms of :

5.2.2

Ask by Summers Chen. in South Africa

Mar 25,2025

Question

QUESTION 3 The roots of a quadratic equation are $$ x=-4 \pm \sqrt{2 p+1} $$. 3.1 Calculate the value(s) of $$ p $$ for which the roots will be real. 3.2 Give one value of $$ p $$ for which the roots will be rational. [3] QUESTION 4 Simplify the following without the use of a calculator: $$ 4.1 \frac{144^{n}\left(3^{-1}ight)^{2 n-1}}{4^{2 n-1}} $$ $$ 4.2 \sqrt[n]{\frac{6^{n}+3^{n+2}}{2^{n}+9}} $$ $$ 4.3 \frac{\sqrt{7}}{\sqrt{63}-7}-\frac{\sqrt{7}}{3 \sqrt{7}+7} $$ [13] QUESTION 5 5.1 If $$ \sqrt{3} \cdot \frac{1}{\cos \alpha}+2=0 $$ and $$ 	an \alpha<0 $$, determine with the aid of a diagram the value of $$ 10 \sin \alpha-3 	an ^{2} \alpha $$. 5.2 If $$ \cos 12^{\circ}=m $$, write down the following in terms of $$ m $$ : $$ 5.2 .1 \sin 168^{\circ} $$ 5.2.2 $$ \cos 192^{\circ} $$ $$ 5.2 .3 	an 372^{\circ} $$

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### QUESTION 3

#### 3.1 Calculate the value(s) of $$ p $$ for which the roots will be real.

The roots of the quadratic equation are given by $$ x = -4 \pm \sqrt{2p + 1} $$. For the roots to be real, the expression under the square root (the discriminant) must be non-negative:

$$
2p + 1 \geq 0
$$

Now, let's solve this inequality for $$ p $$:

$$
2p \geq -1
$$
$$
p \geq -\frac{1}{2}
$$

#### 3.2 Give one value of $$ p $$ for which the roots will be rational.

For the roots to be rational, the expression under the square root must be a perfect square. Let $$ k^2 = 2p + 1 $$ for some integer $$ k $$:

$$
2p = k^2 - 1
$$
$$
p = \frac{k^2 - 1}{2}
$$

Choosing $$ k = 1 $$:

$$
p = \frac{1^2 - 1}{2} = 0
$$

Thus, one value of $$ p $$ for which the roots will be rational is $$ p = 0 $$.

### QUESTION 4

#### 4.1 Simplify $$ \frac{144^{n}(3^{-1})^{2n-1}}{4^{2n-1}} $$

First, let's rewrite the expression:

$$
\frac{144^{n} \cdot 3^{-2n + 1}}{4^{2n - 1}}
$$

Now, we know that $$ 144 = 12^2 = (4 \cdot 3)^2 = 4^2 \cdot 3^2 $$, so:

$$
144^n = (4^2 \cdot 3^2)^n = 4^{2n} \cdot 3^{2n}
$$

Now substituting this back into the expression:

$$
\frac{4^{2n} \cdot 3^{2n} \cdot 3^{-2n + 1}}{4^{2n - 1}} = \frac{4^{2n} \cdot 3^{2n - 2n + 1}}{4^{2n - 1}} = \frac{4^{2n} \cdot 3^{1}}{4^{2n - 1}} = 4^{1} \cdot 3 = 12
$$

#### 4.2 Simplify $$ \sqrt[n]{\frac{6^{n}+3^{n+2}}{2^{n}+9}} $$

We can rewrite $$ 3^{n+2} $$ as $$ 9 \cdot 3^n $$:

$$
\sqrt[n]{\frac{6^n + 9 \cdot 3^n}{2^n + 9}}
$$

Now, $$ 6^n = (2 \cdot 3)^n = 2^n \cdot 3^n $$:

$$
\sqrt[n]{\frac{2^n \cdot 3^n + 9 \cdot 3^n}{2^n + 9}} = \sqrt[n]{\frac{3^n(2^n + 9)}{2^n + 9}} = \sqrt[n]{3^n} = 3
$$

#### 4.3 Simplify $$ \frac{\sqrt{7}}{\sqrt{63}-7}-\frac{\sqrt{7}}{3 \sqrt{7}+7} $$

First, simplify $$ \sqrt{63} = 3\sqrt{7} $$:

$$
\frac{\sqrt{7}}{3\sqrt{7}-7} - \frac{\sqrt{7}}{3\sqrt{7}+7}
$$

Now, find a common denominator:

$$
= \frac{\sqrt{7}(3\sqrt{7}+7) - \sqrt{7}(3\sqrt{7}-7)}{(3\sqrt{7}-7)(3\sqrt{7}+7)}
$$

The numerator simplifies to:

$$
\sqrt{7}(3\sqrt{7}+7 - (3\sqrt{7}-7)) = \sqrt{7}(14) = 14\sqrt{7}
$$

The denominator is:

$$
(3\sqrt{7})^2 - 7^2 = 63 - 49 = 14
$$

Thus, the expression simplifies to:

$$
\frac{14\sqrt{7}}{14} = \sqrt{7}
$$

### QUESTION 5

#### 5.1 If $$ \sqrt{3} \cdot \frac{1}{\cos \alpha}+2=0 $$ and $$ \tan \alpha<0 $$, determine $$ 10 \sin \alpha-3 \tan ^{2} \alpha $$.

From the equation:

$$
\sqrt{3} \cdot \frac{1}{\cos \alpha} = -2 \implies \frac{\sqrt{3}}{\cos \alpha} = -2 \implies \cos \alpha = -\frac{\sqrt{3}}{2}
$$

This corresponds to angles in the second and third quadrants. Since $$ \tan \alpha < 0 $$, we are in the second quadrant.

Thus, $$ \sin \alpha = \frac{1}{2} $$.

Now, calculate $$ 10 \sin \alpha - 3 \tan^2 \alpha $$:

$$
\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} \implies \tan^2 \alpha = \frac{1}{3}
$$

Now substituting:

$$
10 \cdot \frac{1}{2} - 3 \cdot \frac{1}{3} = 5 - 1 = 4
$$

#### 5.2 If $$ \cos 12^{\circ}=m $$, write down the following in terms of $$ m $$:

##### 5.2.1 $$ \sin 168^{\circ} $$

Using the identity $$ \sin(180^\circ - x) = \sin x $$:

$$
\sin 168^{\circ} = \sin(180^{\circ} - 12^{\circ}) = \sin 12^{\circ} = \sqrt{1 - m^2}
$$

##### 5.2.2 $$ \cos 192^{\circ} $$

Using the identity $$ \cos(180^\circ + x) = -\cos x $$:

\[
\cos
### QUESTION 3 1. **Calculate the value(s) of $$ p $$ for which the roots are real:** - For the roots to be real, $$ 2p + 1 \geq 0 $$. - Solving for $$ p $$: $$ p \geq -\frac{1}{2} $$. 2. **Give one value of $$ p $$ for which the roots are rational:** - Choose $$ k = 1 $$: $$ p = \frac{1^2 - 1}{2} = 0 $$. ### QUESTION 4 1. **Simplify $$ \frac{144^{n}(3^{-1})^{2n-1}}{4^{2n-1}} $$:** - Simplifies to $$ 12 $$. 2. **Simplify $$ \sqrt[n]{\frac{6^{n}+3^{n+2}}{2^{n}+9}} $$:** - Simplifies to $$ 3 $$. 3. **Simplify $$ \frac{\sqrt{7}}{\sqrt{63}-7}-\frac{\sqrt{7}}{3 \sqrt{7}+7} $$:** - Simplifies to $$ \sqrt{7} $$. ### QUESTION 5 1. **Determine $$ 10 \sin \alpha - 3 \tan^2 \alpha $$:** - Given $$ \cos \alpha = -\frac{\sqrt{3}}{2} $$ and $$ \sin \alpha = \frac{1}{2} $$. - $$ \tan \alpha = -\frac{1}{\sqrt{3}} $$. - Calculation: $$ 10 \cdot \frac{1}{2} - 3 \cdot \frac{1}{3} = 5 - 1 = 4 $$. 2. **Express the following in terms of $$ m = \cos 12^\circ $$:** - $$ \sin 168^\circ = \sqrt{1 - m^2} $$. - $$ \cos 192^\circ = -\cos 12^\circ = -m $$. - $$ \tan 372^\circ = \tan(360^\circ + 12^\circ) = \tan 12^\circ = \frac{\sin 12^\circ}{\cos 12^\circ} = \frac{\sqrt{1 - m^2}}{m} $$.

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QUESTION 3

QUESTION 4

QUESTION 5

Solución

QUESTION 3

3.1 Calculate the value(s) of for which the roots will be real.

3.2 Give one value of for which the roots will be rational.

QUESTION 4

4.1 Simplify

4.2 Simplify

4.3 Simplify

QUESTION 5

5.1 If and , determine .

5.2 If , write down the following in terms of :

5.2.1

5.2.2

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