Los op vir $x$: 1.1.1 $x^2 - 6x = 0$ 1.1.2 $x^2 + 10x + 8 = 0$ (korrek tot TWEE desimale plekke) 1.1.3 $(1 - x)(x + 2) < 0$ 1.1.4 $ oot{18}{x} - x = 2$ 1.2 Los gelyktydig op vir $x$ en $y$: $x + y = 3$ en $2x^2 + 4xy - y = 15$ 1.3 Indien $n$ die grootste heelgetal is waarvoor $n^{200} < 350^3$, bepaal die waarde van $n$. - NSC Mathematics - Question 1 - 2020 - Paper 1

Using the quadratic formula, we have:

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

where $a = 1$, $b = 10$, and $c = 8$.

Calculating: $x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$ $x = \frac{-10 \pm \sqrt{100 - 32}}{2}$ $x = \frac{-10 \pm \sqrt{68}}{2}$ $x = \frac{-10 \pm 2\sqrt{17}}{2}$ Thus, we get: $x = -5 \pm \sqrt{17}$

Calculating this gives us two values approximately: $x \approx -5 + 4.123 = -0.88$ and $x \approx -5 - 4.123 = -9.12$.

To find the critical points, we set each factor to zero:

$1 - x = 0 \Rightarrow x = 1$ $x + 2 = 0 \Rightarrow x = -2$

We can test intervals: (-∞, -2), (-2, 1), (1, ∞). The sign change will determine where the product is negative. Testing yields:

1. For $x < -2$, both factors are positive.
2. For $-2 < x < 1$, $(1 - x)$ is positive and $(x + 2)$ is positive.
3. For $x > 1$, both factors are negative.

Thus, the solution is: $-2 < x < 1$.

Isolate $x$:

$\sqrt{18}{x} = x + 2$

Squaring both sides yields:

$18x = (x + 2)^2$

Expanding and rearranging gives:

$0 = x^2 - 4x + 4$

Factoring gives: $0 = (x - 2)(x - 2)$

Thus, the solution is $x = 2$.

Starting from the equations:

1. $x + y = 3$
2. $2x^2 + 4xy - y = 15$

From the first equation, solve for $y$:

$y = 3 - x$

Substituting into the second equation:

$2x^2 + 4x(3 - x) - (3 - x) = 15$

This simplifies to:

$2x^2 + 12x - 4x^2 - 3 + x = 15$ $-2x^2 + 13x - 18 = 0$

Factoring gives us: $(x - 2)(x - 9) = 0$

This leads to $x = 2$ or $x = 9$. Using $x + y = 3$, we find corresponding $y$ values: if $x = 2$, $y = 1$, and if $x = 9$, $y = -6$.

To solve for $n$, start by simplifying the inequality:

$n^{200} < 350^3$

Taking the 200th root of both sides gives:

$n < 350^{3/200}$

Calculating: $350^{3/200} = (7 \cdot 50)^{3/200} = n < 11.18$

Since $n$ must be an integer, we conclude:

$n \leq 11$