1.1 Los op vir x: 1.1.1 $x^2 + 5x - 6 = 0$ 1.1.2 $4x^2 + 3x - 5 = 0$ (korrek tot TWEE desimale plekke) 1.1.3 $4x^2 - 1 < 0 $ 1.1.4 $\left( \sqrt{32} + x \right) \left( \sqrt{32} - x \right) = x$ 1.2 Los gelyktydig op vir $x$ en $y$: $y + x = 12$ en $xy = 14 - 3x$ 1.3 Beskryf die produk $1 \times 2 \times 3 \times 4 \times 30.$ Bepaal die grootste waarde van $k$ sodat $3^k$ 'n faktor van die produk is. - NSC Mathematics - Question 1 - 2019 - Paper 1

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 4$, $b = 3$, and $c = -5$.

Substituting gives:

$x = \frac{-3 \pm \sqrt{3^2 - 4(4)(-5)}}{2(4)}$

Calculating the discriminant:

$= \frac{-3 \pm \sqrt{9 + 80}}{8} = \frac{-3 \pm \sqrt{89}}{8}$

Thus, the two solutions are approximately:

1. $x = -1.55$
2. $x = 0.80$

We can rearrange this inequality as:

$4x^2 < 1$

Dividing both sides by 4 results in:

$x^2 < \frac{1}{4}$

Taking the square root gives:

$-\frac{1}{2} < x < \frac{1}{2}$

Expanding the left side gives:

$32 - x^2 = x$

Rearranging leads to:

$x^2 + x - 32 = 0$

Using the quadratic formula:

$x = \frac{-1 \pm \sqrt{1 + 128}}{2} = \frac{-1 \pm 11}{2}$

This results in:

1. $x = 5$
2. $x = -6$. However, the positive solution $x = 4$ can be taken as it fits the inequality from the previous part.

We have two equations:

1. $y + x = 12$ (1)
2. $xy = 14 - 3x$ (2)

Substituting equation (1) into (2) gives:

$x(12 - x) = 14 - 3x$

Simplifying results in:

$x^2 - 15x + 14 = 0$

Factoring gives:

$(x - 14)(x - 1) = 0$

Thus, the solutions for $x$ are:

1. $x = 14 \Rightarrow y = -2$
2. $x = 1 \Rightarrow y = 11$.

To find the largest value of $k$ such that $3^k$ is a factor of the product, we first calculate:

$1 \times 2 \times 3 \times 4 \times 30 = 1 \times 2 \times 3 \times 4 \times (3 \times 10)$

Counting the factors of 3:

• From 3: 1 factor
• From 30: 1 factor

Total factors of 3 in the product: $2$ (i.e., $3^2$). Thus, the largest value of $k$ is $2$.