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

Using the quadratic formula:

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

where $a=3$, $b=-2$, and $c=-6$:

$x = \frac{2 \pm \sqrt{(-2)^{2}-4\cdot3\cdot(-6)}}{2\cdot3}$

Calculating the discriminant:

$= 4 + 72 = 76$

So,

$x = \frac{2 \pm \sqrt{76}}{6} = \frac{2 \pm 2\sqrt{19}}{6} = \frac{1 \pm \sqrt{19}}{3}$

Approximating the values:

• Approximate $x = \frac{1 + \sqrt{19}}{3} \approx 1.12$
• Approximate $x = \frac{1 - \sqrt{19}}{3} \approx -1.79$

Thus, we get:

$x \approx 1.12, \; x \approx -1.79$

To solve this inequality, we first consider where equality holds:

$(x-1)^{2} = 9$

Taking square roots yields:

$x-1 = 3 \quad \text{or} \quad x-1 = -3$

Thus, the critical points are:

$x = 4 \quad \text{and} \quad x = -2$

Now, we test intervals:

• For $x < -2$, choose $x = -3$: $(-3-1)^{2} = 16 > 9$ (True)
• For $-2 < x < 4$, choose $x = 0$: $(0-1)^{2} = 1 < 9$ (False)
• For $x > 4$, choose $x = 5$: $(5-1)^{2} = 16 > 9$ (True)

Thus, the solution set is:

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

To isolate $x$:

$x = 2/\sqrt{6} + 2$

Rationalizing the denominator gives:

$x = \frac{2}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} + 2 = \frac{2\sqrt{6}}{6} + 2 = \frac{\sqrt{6}}{3} + 2$

Thus,

$x = 2 + \frac{\sqrt{6}}{3}$

First, solve $4x+y=2$ for $y$:

$y = 2 - 4x$

Substituting this into the second equation:

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

Expanding this gives:

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

Rearranging gives:

$4x^{2}-12x-4=0$

Dividing further simplifies to:

$x^{2}-3x-1=0$

Using the quadratic formula:

$x = \frac{3 \pm \sqrt{9+4}}{2}=\frac{3 \pm \sqrt{13}}{2}$

Now substituting back to find $y$:

For $x = \frac{3 + \sqrt{13}}{2}$, $y = 2 - 4\left(\frac{3 + \sqrt{13}}{2}\right)$,

And for $x = \frac{3 - \sqrt{13}}{2}$, $y = 2 - 4\left(\frac{3 - \sqrt{13}}{2}\right)$.

We can express 24 as a product of its prime factors:

$24 = 2^{3} \times 3^{1}$

Setting up the equations:

$x = 3\text{ and } y = 1$

Calculating $x - y$ gives:

$x - y = 3 - 1 = 2$