Page 2 of 4 Confidential Information - This is for official consumption Sekhukhune East District Grade 10 Mathematics 11 March 2024 Test 1 QUESTION 1 1.1 Show that the decimal $$ 0, \ddot{2} \dot{8} $$ is a rational number. (3) 1.2 Determine, without the use of a calculator, between which two integers the number $$ \sqrt{42} $$ will lie. 1.3 Factorise the following expressions: 1.3.1 $$ a^{3}-a^{2} $$ 1.3.2 $$ y^{2}-y x-z^{2}-z x $$ 1.4 Expand and simplify: 1.4.1 $$ \frac{y^{2}-y-6}{y^{2}-3 y} \div(y+2) $$ 1.4.2 $$ \frac{3^{2 x+2}-9^{x}}{3^{x} .3^{2 x+1}} $$ 1.5 Determine the value of $$ p $$, if $$ (x-3) $$ is a factor is of $$ 3 x^{2}-p x-6 $$ [18] QUESTION 2 2.1 Determine the value of $$ x $$ if $$ x^{\frac{1}{2}}=5 $$ 2.2 Solve for $$ x $$ : 2.2.1 $$ 2 x-9=3 $$ 2.2.2 $$ x(x-6)=27 $$ $$ 2.2 .3 \quad 2^{x}=\frac{1}{8} $$ 2.2.4 $$ v^{2}=u^{2}+2 a x $$ 2.3 Solve $$ x $$ and $$ y $$ simultaneously: $$ 2 x+3 y=4 \text { and }-6 y-2 x=2 $$

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### QUESTION 1

#### 1.1 Show that the decimal $$ 0, \ddot{2} \dot{8} $$ is a rational number.

To show that $$ 0, \ddot{2} \dot{8} $$ is a rational number, we can express it as a fraction. The notation $$ 0, \ddot{2} \dot{8} $$ means that the digit 2 repeats indefinitely and the digit 8 also repeats indefinitely.

Let $$ x = 0, \ddot{2} \dot{8} $$.

We can express this as:
$$
x = 0.228282828...
$$

To eliminate the repeating decimals, we can multiply $$ x $$ by 1000 (to shift the decimal point three places to the right):
$$
1000x = 228.282828...
$$

Now, we can also multiply $$ x $$ by 10:
$$
10x = 2.282828...
$$

Now, we subtract the second equation from the first:
$$
1000x - 10x = 228.282828... - 2.282828...
$$
$$
990x = 226
$$
$$
x = \frac{226}{990}
$$

Since $$ x $$ can be expressed as a fraction of two integers, it is a rational number.

#### 1.2 Determine, without the use of a calculator, between which two integers the number $$ \sqrt{42} $$ will lie.

To find between which two integers $$ \sqrt{42} $$ lies, we can find the perfect squares around 42.

- $$ 6^2 = 36 $$
- $$ 7^2 = 49 $$

Since $$ 36 < 42 < 49 $$, we have:
$$
6 < \sqrt{42} < 7
$$

Thus, $$ \sqrt{42} $$ lies between the integers 6 and 7.

#### 1.3 Factorise the following expressions:

##### 1.3.1 $$ a^{3}-a^{2} $$

We can factor out the common term $$ a^{2} $$:
$$
a^{3} - a^{2} = a^{2}(a - 1)
$$

##### 1.3.2 $$ y^{2}-y x-z^{2}-z x $$

We can rearrange and group the terms:
$$
y^{2} - yx - z^{2} - zx = (y^{2} - yx) - (z^{2} + zx)
$$
Factoring gives:
$$
y(y - x) - z(z + x)
$$

This can be factored further as:
$$
(y - z)(y + z - x)
$$

#### 1.4 Expand and simplify:

##### 1.4.1 $$ \frac{y^{2}-y-6}{y^{2}-3y} \div(y+2) $$

First, we simplify the division:
$$
\frac{y^{2}-y-6}{y^{2}-3y} \cdot \frac{1}{y+2}
$$

Now, factor the numerator and denominator:
- $$ y^{2} - y - 6 = (y - 3)(y + 2) $$
- $$ y^{2} - 3y = y(y - 3) $$

Now substituting back:
$$
\frac{(y - 3)(y + 2)}{y(y - 3)} \cdot \frac{1}{y + 2}
$$

Cancel $$ (y - 3) $$ and $$ (y + 2) $$:
$$
\frac{1}{y}
$$

##### 1.4.2 $$ \frac{3^{2x+2}-9^{x}}{3^{x} \cdot 3^{2x+1}} $$

We can rewrite $$ 9^{x} $$ as $$ (3^{2})^{x} = 3^{2x} $$:
$$
\frac{3^{2x+2} - 3^{2x}}{3^{x} \cdot 3^{2x+1}} = \frac{3^{2x}(3^{2} - 1)}{3^{x} \cdot 3^{2x+1}} = \frac{3^{2x}(9 - 1)}{3^{x} \cdot 3^{2x+1}} = \frac{3^{2x} \cdot 8}{3^{x} \cdot 3^{2x+1}}
$$

Now simplifying:
$$
= \frac{8}{3^{x} \cdot 3} = \frac{8}{3^{x+1}}
$$

#### 1.5 Determine the value of $$ p $$, if $$ (x-3) $$ is a factor of $$ 3x^{2}-px-6 $$

If $$ (x - 3) $$ is a factor, then substituting $$ x = 3 $$ into the polynomial should yield 0:
$$
3(3)^{2} - p(3) - 6 = 0
$$
$$
27 - 3p - 6 = 0
$$
$$
21 = 3p
$$
$$
p = 7
$$

### QUESTION 2

#### 2.1 Determine the value of $$ x $$ if $$ x^{\frac{1}{2}}=5 $$

Squaring both sides:
$$
x = 5^{2} = 25
$$

#### 2.2 Solve for $$ x $$:

##### 2.2.1 $$ 2x - 9 = 3 $$

Adding 9 to both sides:
$$
2x = 12
$$
Dividing by 2:
$$
x = 6
$$

##### 2.2.2 $$ x(x - 6) = 27 $$

Expanding:
$$
x^{2} - 6x - 27 = 0
$$

Using the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^{2} - 4(1)(-27)}}{2(1)} = \frac{6 \pm \sqrt{36 + 108}}{2} = \frac{6 \pm \sqrt{144}}{2} = \frac{6 \pm 12}{2}
$$

Thus:
$$
x = \frac{18}{2} = 9 \quad \text{or} \quad x = \frac{-6}{2} = -3
$$

##### 2.2.3 $$ 2^{x} = \
### QUESTION 1 1. **1.1** The decimal $$ 0, \ddot{2} \dot{8} $$ is a rational number because it can be expressed as the fraction $$ \frac{226}{990} $$. 2. **1.2** $$ \sqrt{42} $$ lies between the integers 6 and 7. 3. **1.3** - $$ a^{3} - a^{2} = a^{2}(a - 1) $$ - $$ y^{2} - yx - z^{2} - zx = (y - z)(y + z - x) $$ 4. **1.4** - $$ \frac{y^{2} - y - 6}{y^{2} - 3y} \div (y + 2) = \frac{1}{y} $$ - $$ \frac{3^{2x+2} - 9^{x}}{3^{x} \cdot 3^{2x+1}} = \frac{8}{3^{x+1}} $$ 5. **1.5** $$ p = 7 $$ ### QUESTION 2 1. **2.1** $$ x = 25 $$ 2. **2.2** - $$ 2x - 9 = 3 $$ $$ x = 6 $$ - $$ x(x - 6) = 27 $$ $$ x = 9 $$ or $$ x = -3 $$ - $$ 2^{x} = \frac{1}{8} $$ $$ x = -3 $$ - $$ v^{2} = u^{2} + 2a x $$ $$ v = \sqrt{u^{2} + 2a x} $$ - Simultaneous equations: $$ 2x + 3y = 4 \quad \text{and} \quad -6y - 2x = 2 $$ Solving these gives $$ x = 1 $$ and $$ y = \frac{2}{3} $$

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