Los op vir $x$: 1.1.1 $3x^2 + 10x + 6 = 0$ (korrek tot TWEE desimale plekke) 1.1.2 $ ext{ } ext{ } ext{ } ot{\sqrt{6x^2 - 15 = x + 1}}$ 1.1.3 $x^2 + 2x - 24 ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } 0$ Los gelyktydig op vir $x$ en $y$: $5x + y = 3 $ en $3x^2 - 2xy = y^2 - 105$ 1.2.1 Los op vir $p$ as $p^2 - 48p - 49 = 0$ 1.2.2 Vervolgens, of andersins, los op vir $x$ as $7x^2 - 48(7) - 49 = 0$ - NSC Mathematics - Question 1 - 2017 - Paper 1

To solve the equation, we start by isolating the square root

$\sqrt{6x^2 - 15} = x + 1$

Now we square both sides:

$6x^2 - 15 = (x + 1)^2$

This expands to:

$6x^2 - 15 = x^2 + 2x + 1$

Then, rearranging gives:

$5x^2 - 2x - 16 = 0$

Using the quadratic formula again:

$x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 5 \cdot (-16)}}{2 \cdot 5}$

Calculating the discriminant:

$4 + 320 = 324$

Thus,

$x = \frac{2 \pm 18}{10}$

This yields the solutions:

$x_1 = 2, \quad x_2 = -1.6$

To solve this inequality, we start by factoring:

$(x + 6)(x - 4) \geq 0$

The critical points are $x = -6$ and $x = 4$. To determine the intervals, we test values:

• For $x < -6$, say $x = -7$, we find $(-)(-) = +$
• For $-6 < x < 4$, say $x = 0$, we find $(+)(-) = -$
• For $x > 4$, say $x = 5$, we find $(+)(+) = +$

Thus, the solution to the inequality is:

$x \leq -6 \quad \text{or} \quad x \geq 4$

Using the quadratic formula:

$p = \frac{48 \pm \sqrt{(-48)^2 - 4 \cdot 1 \cdot (-49)}}{2 \cdot 1}$

Calculating the discriminant:

$2304 + 196 = 2500$

Thus,

$p = \frac{48 \pm 50}{2}$

Calculating gives:

$p_1 = 49, \quad p_2 = -1$

Simplifying the equation:

$7x^2 - 336 - 49 = 0$

So we have:

$7x^2 - 385 = 0$

Dividing by 7:

$x^2 = \frac{385}{7}$

Taking the square root:

$x = \pm\sqrt{\frac{385}{7}}$

Calculating gives approximate values for $x$.