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 $x^2 + 5x - 4 \\leq 0$ 1.1.4 $\\sqrt{28 - 2 - x} = 2$ 1.2 Los geliktydig vir $x$ en $y$ op in: $2y = 3 + x$ en $2xy + 7 = x^2 + 4y^2$ 1.3 Die wortels van 'n vergelyking is $x = \frac{-b \pm \sqrt{b^2 - 4mp}}{2m}$, waar $m$ en $p$ positiewe reële getalle is - NSC Mathematics - Question 1 - 2021 - Paper 1

Using the quadratic formula:

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

Here, $a = 2$, $b = -3$, and $c = -3$:

$x = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-3)}}{2(2)}$
$= \frac{3 \pm \sqrt{9 + 24}}{4} = \frac{3 \pm \sqrt{33}}{4}$

Calculating the roots:

$x_1 = \frac{3 + \sqrt{33}}{4} \approx 2.19 \\ \\text{and} \\ x_2 = \frac{3 - \sqrt{33}}{4} \approx -0.69$

Rearranging gives:

$x^2 + 5x - 4 = 0$

Finding critical values using the quadratic formula:

$x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-4)}}{2(1)}$
$= \frac{-5 \pm \sqrt{25 + 16}}{2} = \frac{-5 \pm \sqrt{41}}{2}$

The critical points are approximately:

$x = -4 \\text{and} \\ x = -1$

We test intervals to find where the expression is less than zero, which gives:

$x \in [-4, -1]$

Squaring both sides, we have:

$28 - 2 - x = 4$

Simplifying gives:

$26 - x = 4 \\rightarrow x = 22$

Checking for validity:

The expressions under the square root must be non-negative.

From the first equation:

$y = \frac{3 + x}{2}$

Substituting in the second equation:

$2x\left(\frac{3 + x}{2}\right) + 7 = x^2 + 4\left(\frac{3 + x}{2}\right)^2$

Solving this system of equations will yield the values for $x$ and $y$.

Given the quadratic equation structure that incorporates $m$ and $p$, we need to show that the discriminant $\Delta < 0$ leads to non-real roots:

$\Delta = b^2 - 4mp < 0$

This implies that $\Delta < 0$ leads to non-real (imaginary) roots for $x$.