Solve for $x$: 1.1.1 $x (7 + x) = 0$ 1.1.2 $4x^2 - 5x - 4 = 0$ (correct to TWO decimal places) 1.1.3 $2x^2 - 8 > 0$ Solve for $x$ and $y$ if: y = $5x - 2$ and $y = x^2 + 4x - 8$ 1.3 The diagram below shows the movement of a piston inside the engine cylinder of a car - NSC Technical Mathematics - Question 1 - 2022 - Paper 1

Using the quadratic formula, where $a = 4$, $b = -5$, and $c = -4$:

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

Calculating the discriminant:

$b^2 - 4ac = (-5)^2 - 4(4)(-4) = 25 + 64 = 89$

Now substituting back:

$x = \frac{5 \pm \sqrt{89}}{8}$

Calculating the roots:

1. $x_1 = \frac{5 + \sqrt{89}}{8} \approx 1.85$ (to two decimal places)
2. $x_2 = \frac{5 - \sqrt{89}}{8} \approx -0.55$ (to two decimal places).

So, the solutions are approximately $x \approx 1.85$ and $x \approx -0.55$.

Rearranging the inequality:

$2x^2 > 8 \Rightarrow x^2 > 4 \Rightarrow x > 2 \text{ or } x < -2.$

Thus, the solution set is $x > 2$ or $x < -2$.

Equating the two expressions for $y$:

$5x - 2 = x^2 + 4x - 8$

Bringing all terms to one side:

$x^2 - x - 6 = 0$

Factoring:

$(x - 3)(x + 2) = 0$

Thus, $x = 3$ or $x = -2$. Substituting these back:

For $x = 3$: $y = 5(3) - 2 = 13.$ For $x = -2$: $y = 5(-2) - 2 = -12.$

So the solutions are $(3, 13)$ and $(-2, -12)$.

Starting with the formula:

$SV = \frac{\pi d^2 \times L}{4}$

To make $L$ the subject, we multiply both sides by 4 and divide by $\pi d^2$:

$L = \frac{4 \times SV}{\pi d^2}$.

Substituting into the formula:

$L = \frac{4 \times 1,020.5}{\pi (10)^2} = \frac{4 \times 1,020.5}{\pi \times 100}$

Calculating:

$L \approx \frac{4,082}{314.16} \approx 13 \text{ cm}.$

$P = 1010_2 = 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 0 \times 2^0 = 10$.

$Q = 1000_2 = 8$.

Thus, $P \times Q = 10 \times 8 = 80$. In binary form, $80_{10} = 1010000_2$.