Solve for $x$: 1.1.1 $x^2 - 2x - 24 = 0$ 1.1.2 $2x^2 - 3x - 3 = 0$ (correct to TWO decimal places) 1.1.3 $x^2 + 5x < -4$ 1.1.4 $\sqrt{28} = 2 - x$ Solve simultaneously for $x$ and $y$ in: 2y = 3 + x and 2xy + 7 = x^2 + 4y^2 1.3 The roots of an equation are $x = \frac{-n \pm \sqrt{n^2 - 4mp}}{2m}$ where $m$, $n$, and $p$ are positive real numbers - NSC Mathematics - Question 1 - 2021 - Paper 1

We can use the quadratic formula to find the roots of the equation:

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

Here, $a = 2$, $b = -3$, and $c = -3$. Substituting these values in gives:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2}$

This simplifies to:

$x = \frac{3 \pm \sqrt{9 + 24}}{4} = \frac{3 \pm \sqrt{33}}{4}$

Calculating the two decimal values:

$x \approx 2.19\ \ ext{ and } \ x \approx -0.69.$

Rearranging the inequality gives:

$x^2 + 5x + 4 < 0$

Next, we can factor this to find critical values:

$(x + 4)(x + 1) < 0$

The critical points are $x = -4$ and $x = -1$. To determine the intervals, we can test values from the intervals $(-\infty, -4)$, $(-4, -1)$, and $(-1, \infty)$:

1. For $x < -4$, the expression is positive.
2. For $-4 < x < -1$, the expression is negative.
3. For $x > -1$, the expression is positive.

Thus, the solution is:

$x \in (-4, -1).$

We'll start by squaring both sides to eliminate the square root:

$28 = (2 - x)^2$

Expanding gives:

$28 = 4 - 4x + x^2$

Rearranging leads to:

$x^2 - 4x - 24 = 0$

Factoring gives:

$(x - 6)(x + 4) = 0$

Thus, $x = 6$ or $x = -4$. After evaluating, we find:

$x = 8 \text{ or } x = -3.$

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

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

Substituting this into the second equation:

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

Expanding and simplifying gives:

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

From here, we can solve for $x$ and $y$ in terms of one another.

Given that $m$, $n$, and $p$ form a geometric sequence, we can express this as:

$n^2 = mp, \, \text{and } p = \frac{n^2}{m}$

Substituting these into the root equation yields:

$\Delta < 0 \Rightarrow n^2 - 4mp < 0 \implies \text{which holds if } m, n, p \text{ are positive real.}$

Therefore, we conclude that the roots $x$ are non-real, as required.