Los op vir $x$: 1.1.1 \( (3x - 1)(x + 4) = 0 \) 1.1.2 \( 2x^2 + 9x - 14 = 0 \) (korrek tot TWEE desimale plekke) 1.1.3 \( \sqrt{3 - 26x} = 3x \) 1.1.4 \( (x - 1)(x - 4) = x + 11 \) 1.2 Vereenvoudig volledig: \( \frac{\sqrt{16x^2} - \sqrt{25x^2}}{\sqrt{x}} \) 1.3 Los gelyktydig op vir $x$ en $y$: $xy = 9$ en $-2y - 3 = 0$ 1.4 Bewys dat \( x^2 + 2xy + y^2 \) nie negatief vir $x, y \in \mathbb{R}$ 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, c = -14 ):

1. Calculate the discriminant: [ b^2 - 4ac = 9^2 - 4(2)(-14) = 81 + 112 = 193 ]

2. Apply the quadratic formula: [ x = \frac{-9 \pm \sqrt{193}}{4} ]

Thus, the approximate solutions are ( x \approx 1.22 ) and ( x \approx -5.72 ).

To solve ( \sqrt{3 - 26x} = 3x ), square both sides:

[ 3 - 26x = 9x^2 ]

Rearranging gives:

[ 9x^2 + 26x - 3 = 0 ]

Using the quadratic formula: [ x = \frac{-26 \pm \sqrt{26^2 - 4(9)(-3)}}{2(9)} ]

Solve for the roots.

To solve ( (x - 1)(x - 4) = x + 11 ), expand the left side: [ x^2 - 5x + 4 = x + 11 ]

Rearranging gives: [ x^2 - 6x - 7 = 0 ]

Using the quadratic formula: [ x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(-7)}}{2(1)} ]

Calculate roots.

Simplifying gives: [ \frac{4x - 5x}{\sqrt{x}} = \frac{-x}{\sqrt{x}} = -\sqrt{x} ]

From the second equation, solve for $y$: [ -2y = 3 \rightarrow y = -\frac{3}{2} ]

Substitute $y$ into $xy = 9$: [ x(-\frac{3}{2}) = 9 \rightarrow x = -6 ] Thus, $x = -6$ and $y = -\frac{3}{2}$.

We know:[ x^2 \geq 0 \quad y^2 \geq 0 \quad \text{and} \quad 2xy \geq 0 \text{ when } xy \geq 0 ]

Thus, ( x^2 + 2xy + y^2 = (x+y)^2 \geq 0 ) for all real $x$ and $y$.