1.1 Los op vir $x$: 1.1.1 $x^2 - 2x - 24 = 0$ 1.1.2 $2x^2 - 3x - 3 = 0$ (korrek tot TWEE desimale syfers) 1.1.3 $x^2 + 5x ext{ ≤ } 4$ 1.1.4 $\sqrt{28 - 2 - x} = 2$ 1.2 Los gelijktydig vir $x$ en $y$ op in: $2y = 3 + x$ en $2y + 7 = x^2 + 4y^2$ 1.3 Die wortels van 'n vergelyking is $y = \frac{-n \pm \sqrt{n^2 - 4mp}}{2m}$, waar $m$ en $p$ positiewe reële getalle is - NSC Mathematics - Question 1 - 2021 - Paper 1

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 2$, $b = -3$, and $c = -3$. Substituting:

= \frac{3 \pm \sqrt{9 + 24}}{4} = \frac{3 \pm \sqrt{33}}{4}$$ Calculating the two solutions: $$x_1 = \frac{3 + \sqrt{33}}{4} \approx 2.19 \text{and} x_2 = \frac{3 - \sqrt{33}}{4} \approx -0.69$$.

Rearranging the inequality gives:

$x^2 + 5x - 4 \leq 0$
Next, we factor or use the quadratic formula to find the critical points:

= \frac{-5 \pm \sqrt{25 + 16}}{2} = \frac{-5 \pm \sqrt{41}}{2}$$ The critical values are approximately: $$x_1 \approx -4 \text{ and } x_2 \approx -1$$ Thus, the solution set is: $$x \in [-4, -1]$$.

First, we square both sides:

$28 - 2 - x = 4$
Rearranging gives:

$x = 28 - 2 - 4 = 22$
Thus, the solution is:

$x = 22$.

From the first equation:

$2y = 3 + x \Rightarrow y = \frac{3 + x}{2}$
Substituting into the second equation:

$2\left(\frac{3 + x}{2}\right) + 7 = x^2 + 4\left(\frac{3 + x}{2}\right)^2$
This leads to a combined equation where $x$ and $y$ can be solved simultaneously.

Given the roots of the equation:

$y = \frac{-n \pm \sqrt{n^2 - 4mp}}{2m}$
Since both $m$ and $p$ are positive, we know that $n^2 - 4mp < 0$ results in non-real roots. Hence:

$\Delta < 0 \Rightarrow x \text{ is a non-real number.}$