Los op vir $x$: 1.1.1 $(3x - 6)(x + 2) = 0$ 1.1.2 $2x^2 - 6x + 1 = 0$ (korrek tot TWEVE desimale plekke) 1.1.3 $x^2 - 90 > x$ 1.1.4 $x - 7/ extstyle{ oot{x}} = -12$ Los gelktydig vir $x$ en $y$ op: $2x - y = 2$ $xy = 4$ - NSC Mathematics - Question 1 - 2022 - Paper 1

To solve $2x^2 - 6x + 1 = 0$, we apply the quadratic formula:

x = rac{-b ext{ ± } ext{√}(b^2 - 4ac)}{2a} Where $a = 2$, $b = -6$, and $c = 1$.

Calculating $b^2 - 4ac$: $b^2 - 4ac = (-6)^2 - 4(2)(1) = 36 - 8 = 28$

Now substituting into the quadratic formula: x = rac{6 ext{ ± } ext{√}28}{4} = rac{6 ext{ ± } 2 ext{√}7}{4} = rac{3 ext{ ± } ext{√}7}{2}

So the solutions are: $x ext{ } ≈ 2.82 ext{ and } 0.18$.

Rearranging the inequality gives: $x^2 - x - 90 > 0$ Factoring or using the quadratic formula can find the critical values: Calculate using the quadratic formula, where $a = 1$, $b = -1$, and $c = -90$:

x = rac{-(-1) ext{ ± } ext{√}((-1)^2 - 4(1)(-90))}{2(1)}

This leads to critical values: $x = 10 ext{ and } -9$

The intervals to test are $(- extstyle{ ext{-∞}}, -9)$, $(-9, 10)$, and $(10, ext{ +∞})$. Testing these intervals shows: The solution is: $x < -9 ext{ or } x > 10.$

Isolate the root:

oot{x}} + 12 = 0$$ Rearranging gives: $$ extstyle{ oot{x}} = 3 ext{ or } extstyle{ oot{x}} = 4$$ Squaring both sides: $$x = 9 ext{ or } x = 16$$ Therefore, the solutions are: $$x = 9 ext{ and } x = 16.$$

We have the equations:

1. $2x - y = 2$
2. $xy = 4$

From the first equation, express $y$ in terms of $x$: $y = 2x - 2$

Substituting into the second equation: $x(2x - 2) = 4$ Expanding and rearranging gives: $2x^2 - 2x - 4 = 0$

dividing by 2: $x^2 - x - 2 = 0$ Factoring gives: $(x - 2)(x + 1) = 0$ The solutions for $x$ are: $x = 2 ext{ or } x = -1$

Now substituting back to find $y$:

• For $x = 2$: $y = 2(2) - 2 = 2$
• For $x = -1$: $y = 2(-1) - 2 = -4$ Thus the pairs $(x, y)$ are $(2, 2)$ and $(-1, -4)$.