1.1 Los op vir x: 1.1.1 $x^2 + 2x - 15 = 0$ (3) 1.1.2 $5x^2 - x - 9 = 0$ (Los jou antwoord korrek tot TWEE desimale syfers.) (4) 1.1.3 $x^2 \\leq 3x$ (4) 1.2 Gegee: $\frac{a + 64}{a} = 16$ (3) 1.2.1 Los op vir a - NSC Mathematics - Question 1 - 2022 - Paper 1

For the equation $5x^2 - x - 9 = 0$, we apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 5$, $b = -1$, and $c = -9$. Let's calculate:

1. Calculate the discriminant: $b^2 - 4ac = (-1)^2 - 4(5)(-9) = 1 + 180 = 181$

2. Plug the values back into the formula: $x = \frac{-(-1) \pm \sqrt{181}}{2(5)} = \frac{1 \pm \sqrt{181}}{10}$

Approximating the square root:

• $\sqrt{181} \approx 13.45$

Thus, the solutions are: $x \approx \frac{1 + 13.45}{10} \approx 1.45 \quad \text{and} \quad x \approx \frac{1 - 13.45}{10} \approx -1.25$

To solve $x^2 \leq 3x$, we first rearrange it to standard form:

$x^2 - 3x \leq 0$

Factoring gives: $(x)(x - 3) \leq 0$

Next, we find the critical points, which are $x = 0$ and $x = 3$. Testing intervals:

• For $x < 0$, choose $x = -1$: $(-1)(-4) = 4$ (not in solution)
• For $0 < x < 3$, choose $x = 1$: $(1)(-2) = -2$ (in solution)
• For $x > 3$, choose $x = 4$: $(4)(1) = 4$ (not in solution)

The solution is therefore: $0 \leq x \leq 3$

We start with the equation: $\frac{a + 64}{a} = 16$

Cross-multiplying yields: $a + 64 = 16a$

Rearranging to isolate $a$: $64 = 15a \Rightarrow a = \frac{64}{15} \approx 4.27$

Starting with the equation: $2^x + 2^{6+x} = 16$

We can express 16 as $2^4$, leading to: $2^x + 2^6 \cdot 2^x = 2^4$

Factoring gives: $2^x(1 + 64) = 2^4 \Rightarrow 65 imes 2^x = 2^4$

Dividing by 65: $2^x = \frac{2^4}{65}$

Taking log base 2: $x = \log_2 \left( \frac{16}{65} \right)$

We can factor the expression inside the square root:

1. Rewrite the numerator: $2^{1002} + 2^{1006} = 2^{1002}(1 + 2^4) = 2^{1002} \cdot 17$

2. The expression then simplifies: $\sqrt{\frac{2^{1002} \cdot 17}{17(2)^{99}}} = \sqrt{\frac{2^{1002}}{(2)^{99}}} = \sqrt{2^{3}} = 2^{1.5} = 2 \sqrt{2}$

We have the system of equations:

1. $2x - y = 2$ (1)
2. $\frac{1}{x} - 3y = 1$ (2)

From (1), express $y$: $y = 2x - 2$

Substituting into (2): $\frac{1}{x} - 3(2x - 2) = 1$ $\frac{1}{x} - 6x + 6 = 1$ $\frac{1}{x} = 6x - 5$ Multiplying through by $x$ gives: $1 = 6x^2 - 5x$ $6x^2 - 5x - 1 = 0$ Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Where $a = 6$, $b = -5$, and $c = -1$. The roots can be calculated to find corresponding values of $y$.