Los op vir $x$: 1.1.1 $(x–3)(x+1)=0$ 1.1.2 $ ext{sqrt}{x}=512$ 1.1.3 $x(x–4) < 0$ Gegee: 1.2 $f(x) = x^2 - 5x + 2$ 1.2.1 Los op vir $x$ as $f(x) = 0$ 1.2.2 Vir watter waardes van $c$ sal $f(x) = c$ geen reële wortels hê nie? 1.3 Los op vir $x$ en $y$: $x = 2y + 2$ $x^2 - 2xy + 3y^2 = 4$ 1.4 Bereken die maksimum waarde van $S$ as $S = rac{6}{x^2 + 2}$ - NSC Mathematics - Question 1 - 2017 - Paper 1

To isolate $x$, we first square both sides of the equation:

$\text{sqrt}{x}=512 \Rightarrow x = 512^2$

Calculating gives:

$x = 262144$

Thus, the solution is $x = 64$.

To solve the inequality $x(x–4) < 0$, we find the critical values by solving $x(x–4) = 0$:

• $x = 0$
• $x = 4$

Next, we test intervals between these critical values:

• For $x < 0$, choose $x = -1$: $(-1)(-1-4) = 5 > 0$
• For $0 < x < 4$, choose $x = 2$: $(2)(2-4) = -4 < 0$
• For $x > 4$, choose $x = 5$: $(5)(5-4) = 5 > 0$

Thus, the solution for the inequality is $0 < x < 4$, or $x \in (0, 4)$.

To find the roots of the function $f(x) = x^2 - 5x + 2$, we set it to zero:

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

Using the quadratic formula:

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

after substituting $a=1, b=-5, c=2$ gives:

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

which simplifies to:

$x = \frac{5 \pm \sqrt{25 - 8}}{2} = \frac{5 \pm \sqrt{17}}{2}$

Calculating the values gives:

$x \approx 0.44 \text{ or } x \approx 4.56$.

To determine when the function $f(x) = x^2 - 5x + 2$ has no real roots, we analyze the discriminant given by $D = b^2 - 4ac$:

• For no real roots, $D < 0$. Here, that means:

$(-5)^2 - 4(1)(c) < 0 \Rightarrow 25 - 4c < 0$

Solving gives:

$25 < 4c \Rightarrow c > \frac{25}{4} = 6.25$

Therefore, $f(x) = c$ has no real roots for $c > 6.25$.

First, substitute $x$ in the second equation:

$x^2 - 2xy + 3y^2 = 4$

Replacing $x$ gives:

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

Expanding yields:

$4y^2 + 8y + 4 - (4y^2 + 4y) + 3y^2 = 4$

Combine like terms:

$3y^2 + 4y + 4 - 4 = 0 \Rightarrow 3y^2 + 4y = 0$

Factoring gives:

$y(3y + 4) = 0 \Rightarrow y = 0 \text{ or } y = -\frac{4}{3}$

Thus, with $y = 0$, $x = 2(0) + 2 = 2$. For $y = -\frac{4}{3}$:

i gives: $x = 2\left(-\frac{4}{3}\right) + 2 = -\frac{8}{3} + 2 = \frac{-2}{3}$

Therefore, the pairs are $(2, 0)$ and $(\frac{-2}{3}, -\frac{4}{3})$.

To find the maximum value of the function $S = \frac{6}{x^2 + 2}$, we take the derivative and set it to zero:

$S' = -\frac{12x}{(x^2 + 2)^2}$

Setting $S' = 0$ results in:

$-12x = 0 \Rightarrow x = 0$.

To find the maximum, we evaluate $S$ at critical points:

$S(0) = \frac{6}{0^2 + 2} = 3$

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