1.1 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 $\sqrt{28} = 2 - x$ 1.2 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 solve this using the quadratic formula:

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

Where $a = 2$, $b = -3$, and $c = -3$. Substituting these values:

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

Calculating the two possible values of $x$ gives approximately:

• $x \approx 2.19$
• $x \approx -0.69$

First, we rewrite the inequality in standard form:

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

Now we factor the left-hand side:

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

Next, we find the critical points where the expression equals zero, which are $x = -4$ and $x = -1$. To solve the inequality, we test intervals around these points:

1. For $x < -4$, both factors are negative, the product is positive.
2. For $-4 < x < -1$, the first factor is positive and second is negative, the product is negative.
3. For $x > -1$, both factors are positive, the product is positive.

Thus, the solution to the inequality is:

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

First, we square both sides to eliminate the square root:

$28 = (2 - x)^2$

Expanding the right side gives:

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

Rearranging leads to:

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

We factor this equation to find: $(x - 6)(x + 4) = 0$

The possible solutions are:

1. $x = 6$
2. $x = -4$

From the first equation, we solve for $y$:

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

Substituting this into the second equation, we get:

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

We can simplify and solve this equation for $x$ and then find the corresponding $y$. This involves expansion and solving a quadratic equation.

Given that $m$, $n$, and $p$ form a geometric sequence, we have:

$n^2 = mp$

Using the formula for $x$, substituting $n$ gives:

$x = \frac{-n \pm \sqrt{n^2 - 4mp}}{2m} = \frac{-n \pm \sqrt{n^2 - 4n^2}}{2m} = \frac{-n \pm \sqrt{-3n^2}}{2m}$

Since the term under the square root is negative, $x$ will have non-real values, proving that $x$ is a non-real number.