Los op vir $x$: 1.1.1 $(3x-1)(x+4) = 0$ 1.1.2 $2x^2 + 9x - 14 = 0$ (korrek tot TWEE decimal plekke) 1.1.3 $rac{ ext{√}3 - 26}{x} = 3$ 1.1.4 $(x-1)(x-4) = x + 11$ 1.2 Vereenidig volledig: $rac{ ext{√}16x^2 - ext{√}25x^2}{ ext{√}x}$ 1.3 Los geliktydig op vir $x$ en $y$: $xy = 9$ en $-2y -3 = 0$ 1.4 Bewys dat $x^2 + 2xy + 2y^2$ nie negatief vir $x, y ext{ in } R$ kan wees nie. - NSC Mathematics - Question 1 - 2018 - Paper 1

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 2$, $b = 9$, and $c = -14$.

Substituting these values:

$x = \frac{-9 \pm \sqrt{9^2 - 4(2)(-14)}}{2(2)}$

Calculate the discriminant:

$= 81 + 112 = 193$

So we have:

$x = \frac{-9 \pm \sqrt{193}}{4}$

Calculating values:

1. $x = \frac{-9 + \sqrt{193}}{4} \approx 1,22$.
2. $x = \frac{-9 - \sqrt{193}}{4} \approx -5,72$.

Multiply both sides by $x$: $\text{√}3 - 26 = 3x$

Thus, rearranging gives: $x = \frac{\text{√}3 - 26}{3}$.

First, expand the left side:

$(x-1)(x-4) = x^2 - 4x - x + 4 \\ = x^2 - 5x + 4$

Set the equation to zero: $x^2 - 5x + 4 - x - 11 = 0 \\ = x^2 - 6x - 7 = 0$

Now, factor the quadratic: $(x-7)(x+1) = 0$ Thus, solutions are $x = 7$ and $x = -1$.

First, simplify the expression:

Substituting values, we have: $\text{√}16x^2 = 4x \\ \text{and } \text{√}25x^2 = 5x$ Thus: $= \frac{4x - 5x}{\text{√}x} = \frac{-x}{\text{√}x} = -\text{√}x$.

From $-2y - 3 = 0$, we can express $y$:

1. $y = -\frac{3}{2}$.

Now, substitute $y$ into $xy = 9$: $x(-\frac{3}{2}) = 9 \\ = x = -6 ext{ or } y = -3$.

The expression $x^2 + 2xy + 2y^2$ can be rearranged.

This can be represented as non-negative since:

1. $x^2 \ge 0$.
2. $y^2 \ge 0$.
3. Therefore, $(x + y)^2 \ge 0$.

Thus it shows that $x^2 + 2xy + 2y^2 \ge 0$.