1.1 Los op vir $x$: 1.1.1 $x^2 - 4x + 3 = 0$ 1.1.2 $5x^2 - 5x + 1 = 0$ (korrek tot TWEE desimale plekke) 1.1.3 $x^2 - 3x - 10 > 0$ 1.1.4 $3 ad{x} = x - 4$ 1.2 Los gelijktijdig op vir $x$ en $y$: $3x - y = 2$ en $2y + 9x^2 = -1$ 1.3 Indien $3^x = 64$ en $5^{rac{1}{ ad{y}}} = 64$, SONDERS die gebruik van 'n sakrekenaar, die waarde van: $$rac{3^{rac{1}{2}}}{ ad{5^y}}$$ - NSC Mathematics - Question 1 - 2018 - Paper 1

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 5$, $b = -5$, and $c = 1$:

Calculate the discriminant: $b^2 - 4ac = (-5)^2 - 4(5)(1) = 25 - 20 = 5$ Then, $x = \frac{5 \pm \sqrt{5}}{10} = \frac{1}{2} \pm \frac{\sqrt{5}}{10}$ Calculating the values gives us: $x \approx 0,72$ and $x \approx 0,28$.

To solve this inequality, first factor:
$(x - 5)(x + 2) > 0$ Critical points are $x = -2$ and $x = 5$.
We analyze intervals:

1. $(-\infty, -2)$
2. $(-2, 5)$
3. $(5, \infty)$
   Evaluating these intervals gives: $x < -2 \text{ or } x > 5$

Squaring both sides gives: $9x = (x - 4)^2$ Expanding the right side: $9x = x^2 - 8x + 16$ Rearranging results in: $x^2 - 17x + 16 = 0$ Factoring leads to the critical points: $(x - 16)(x - 1) = 0$ Thus, $x = 16$ or $x = 1$.

From the first equation, we can express $y$ in terms of $x$: $y = 3x - 2$ Substituting this in the second equation: $2(3x - 2) + 9x^2 = -1$ $6x - 4 + 9x^2 = -1$ Rearranging gives: $9x^2 + 6x - 3 = 0$ Dividing through by 3: $3x^2 + 2x - 1 = 0$ Factoring gives the solutions for $x$:
$x = \frac{1}{3}$ or $x = -1$.
Substituting these back: $y = 3(\frac{1}{3}) - 2 = -1 \text{ and } y = 3(-1) - 2 = -5$ So, $(x, y) = (\frac{1}{3}, -1)$ and $(-1, -5)$.

First, solve $3^x = 64$:
Since $64 = 4^3$, we have: $3^x = 4^3$ Taking logarithms: $x = \frac{3 \log(4)}{\log(3)}$ Now for $5^{\frac{1}{\rad{y}}} = 64$:
$\frac{1}{\rad{y}} = 3 \implies y = 9$ Thus, substituting into the equation: $\frac{3^{\frac{1}{2}}}{\rad{5^9}} = \frac{\sqrt{3}}{5^{4.5}}$