Los op vir $x$: 1.1.1 $x^2 - x - 20 = 0$ 1.1.2 $3x^2 - 2x - 6 = 0$ (korrek tot TWEE desimale sifers) 1.1.3 $(x - 1)^2 > 9$ 1.1.4 $2/ oot{x}{6} + 2 = x$ Los gelyktydig op vir $x$ en $y$: 1.2 $4x + y = 2$ en $4x + y^2 = 8$ Indien dit gegee word dat $2^x imes 3^y = 24^2$, bepaal die numeriese waarde van $x - y$. - NSC Mathematics - Question 1 - 2021 - Paper 1

For this equation, we will use the quadratic formula:

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

Here, $a = 3$, $b = -2$, and $c = -6$.

So,

$x = \frac{2 \pm \sqrt{(-2)^2 - 4(3)(-6)}}{2(3)}$ $x = \frac{2 \pm \sqrt{4 + 72}}{6}$ $x = \frac{2 \pm \sqrt{76}}{6}$ $x = \frac{2 \pm 2\sqrt{19}}{6}$ $x = \frac{1 \pm \sqrt{19}}{3}$

The two decimal approximations yield:

$x \approx 1.12 \quad \text{and} \quad x \approx -1.79.$

To solve this inequality, we start by rewriting it:

$(x - 1)^2 > 9$

Taking the square root of both sides, we have two cases:

$x - 1 > 3 \Rightarrow x > 4$ $x - 1 < -3 \Rightarrow x < -2$

Thus, the solution to this inequality is:

$x < -2 \quad \text{or} \quad x > 4$

First, we isolate the surd:

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

Rearranging this gives:

$\frac{2}{\sqrt{x}} = x - 2$

Now, cross-multiplying leads to:

$2 = (x - 2) \sqrt{x}$

Squaring both sides results in:

$4 = (x - 2)^2 x$

Expanding and rearranging will yield the solutions for $x$.

From the first equation, we can express $y$ in terms of $x$:

$y = 2 - 4x$

Substituting this into the second equation gives:

$4x + (2 - 4x)^2 = 8$

Expanding the quadratic results in:

$4x + (4 - 16x + 16x^2) = 8$

Simplifying this leads to a quadratic equation in terms of $x$. Solve for $x$ to find the corresponding $y$.

We can express $24$ as $24 = 2^3 imes 3^1$, so:

$24^2 = (2^3 imes 3^1)^2 = 2^6 imes 3^2$

Equating the exponents gives:

$x = 6 \text{ and } y = 2$

Finally, we find:

$x - y = 6 - 2 = 4.$