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 $3rac{ oot{}}{x} = 4$ Los geliktydig op vir $x$ en $y$: 3$x - y = 2$ en 2$y + 9$x$^2 = -1$ Indien $3^x = 64$ van $5rac{ oot{}}{p} = 64$, SONDERS die gebruik van 'n sakrekenaar, die waarde van: $rac{[3^x]}{ oot{5}{p}}$ - NSC Mathematics - Question 1 - 2018 - Paper 1

Using the quadratic formula:

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

For our equation, where $a = 5$, $b = -5$, and $c = 1$:

1. Calculate the discriminant: $b^2 - 4ac = (-5)^2 - 4(5)(1) = 25 - 20 = 5$
2. Substitute into the formula: $x = \frac{5 \pm \sqrt{5}}{10}$
3. This gives:

• $x \approx 0.72$
• $x \approx 0.28$

Factoring the quadratic we find:

$(x - 5)(x + 2) > 0$

This means:

• $x < -2$ or $x > 5$

Squaring both sides gives:

$3x = 16$

Thus:

$x = \frac{16}{3}$

From the first equation, we find: $y = 3x - 2$

Substituting into the second equation:

$2(3x - 2) + 9x^2 = -1$

This simplifies to: $18x - 4 + 9x^2 = -1$

Rearranging gives: $9x^2 + 18x - 3 = 0$

Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Solving this gives the $x$ values. Substitute back to find $y$ values.

Since $3^x = 64$, we recognize: $4^3 = 64 \Rightarrow 3^x = 4^3$ Thus, $x = 4$. Next, if $\sqrt{p} = 64$, then $p = 8$. To calculate: $\frac{3^x}{\sqrt{5}{p}} = \frac{64}{8} = 8$