Los op vir $x$: 1.1.1 $3x^2 + 10x + 6 = 0$ (korrek tot TWEE desimale plekke) 1.1.2 $ ewline ext{Let: } ewlineegin{align*} ext{Let } z = rac{ ext{ } ext{Given}}{ ext{ } ext{formula}}\ ext{With:} ewline ext{ } ext{the expression} ewline ext{ } ext{is} ext{allocated × } 3 \ ext{ } = ewline ext{} ext{Answer is } ewline ext{} ext{when calculated.}\ ext{ } ext{answer is with ext{ expressions which follow as} }\ ext{also for the next} ewline ext{2.4 } ext{solved as follow a similar method,}\ ext{Answer needed also for the other expressions that follow } ext{and such that:}\ ext{the steps will lead to the same - English General - NSC Mathematics - Question 1 - 2017 - Paper 1

The next task involves simplifying the equation $\sqrt{6x^2 - 15} = x + 1$. Begin by squaring both sides:

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

Expanding the right side results in:

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

Rearranging gives:

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

Applying the quadratic formula again:

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

Calculating:

$x = \frac{2 \pm \sqrt{4 + 320}}{10} = \frac{2 \pm \sqrt{324}}{10} = \frac{2 \pm 18}{10}$

This results in:

• $x = 2$ or $x = -1.6$

To solve the inequality $x^2 + 2x - 24 \geq 0$, first find the roots of the equation by factoring:

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

The critical points are $x = -6$ and $x = 4$. The intervals created are:

1. $(-\infty, -6)$
2. $[-6, 4]$
3. $(4, +\infty)$

Testing a point from each interval:

• For $x < -6$, choose $x = -7$, which gives a positive value.
• For $-6 < x < 4$, choose $x = 0$, which gives a negative value.
• For $x > 4$, choose $x = 5$, which gives a positive value.

Thus, the solution is:

$x \in (-\infty, -6] \cup [4, +\infty)$