Los op vir $x$: 1.1.1 $(x-3)(x+1) = 0$ 1.1.2 $\\sqrt{x} = 512$ 1.1.3 $x(x-4) < 0$ Gegee: $f(x) = x^2 - 5x + 2$ 1.2.1 Los op vir $x$ as $f(x) = 0$ 1.2.2 Vir watter waardes van $c$ sal $f(c) = c$ geen reële wortels hê nie? Los op vir $x$ en $y$: $x = 2y + 2$ $x^2 - 2xy + 3y^2 = 4$ 1.4 Bereken die maksimum waarde van $S$ as $S = \frac{6}{x^2 + 2}$. - NSC Mathematics - Question 1 - 2017 - Paper 1

To solve for $x$, we square both sides:

$x = (512)^2$

Calculating, we find:

$x = 262144$

Therefore, $x = 64$.

To solve the inequality, we first find the critical values by setting the equation to zero:

1. $x(x - 4) = 0$ gives us $x = 0$ and $x = 4$.

Next, we test intervals to determine where the product is negative:

• For $x < 0$, both factors are negative, hence positive product.
• For $0 < x < 4$, the product is negative.
• For $x > 4$, both factors are positive, hence positive product.

Thus, the solution is $0 < x < 4$.

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = -5$, and $c = 2$:

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

Calculating the discriminant: $b^2 - 4ac = 25 - 8 = 17$

Thus, the solutions are:

$x = \frac{5 \pm \sqrt{17}}{2}$ This gives approximate values:

1. $x \approx 4.56$
2. $x \approx 0.44$.

To find the values of $c$ such that $f(c) - c = 0$ leads to no real roots:

$c^2 - 5c + 2 - c = 0\Rightarrow c^2 - 6c + 2 = 0$ Using the discriminant: $D = b^2 - 4ac = (-6)^2 - 4(1)(2) = 36 - 8 = 28 > 0.$

Thus, there are always roots for all values of $c$. Therefore, the question becomes irrelevant when analyzing the discriminant, ensuring that for certain conditions on the coefficients, $c^2 - 6c + 2$ should hold negative discriminant ensuring no real roots.

Substituting $x = 2y + 2$ into the second equation:

$(2y + 2)^2 - 2(2y + 2)y + 3y^2 = 4$

Expanding the equation:

$4y^2 + 8y + 4 - (4y^2 + 4y + 6y^2) =4$

This simplifies to: $(1y^2 - 3y + 4 = 4)\Rightarrow y^2 - 3y = 0\Rightarrow y(y - 3) = 0\Rightarrow y = 0 or y = 3$

Now substituting back to find $x$:

1. When $y = 0$, $x = 2(0) + 2 = 2$.
2. When $y = 3$, $x = 2(3) + 2 = 8$.

Thus, $(x,y)$ pairs are $(2,0)$ and $(8,3)$.

To find the maximum of $S$, we rewrite it:

$S = 6(x^2 + 2)^{-1}$

To maximize $S$, we minimize $x^2 + 2$. Since $x^2 + 2 \geq 2$ for all $x$, then the minimum occurs at $x^2=0$. Thus, the maximum value of:

$S_{max} = \frac{6}{2} = 3.$

Therefore, the maximum value of $S$ is 3.