1.1 Los op vir x: 1.1.1 $x^2 + x - 12 = 0$ 1.1.2 $3x^2 - 2x = 6$ (antwoorde korrekt tot TWEE desimale plekke) 1.1.3 $ ext{√}(2x+1) = x - 1$ 1.1.4 $x^2 - 3 > 2x$ 1.2 Los gelyktydig vir x en y op: $x + 2 = 2y$ $rac{1}{x} + y = 1$ 1.3 Gegee: $2^{m+4} + 2^{n} = 3^{n-2} - 3$ waar m en n heelgetalle is - NSC Mathematics - Question 1 - 2023 - Paper 1

First, bring all terms to one side to set the equation to zero:

$3x^2 - 2x - 6 = 0$

Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, with $a = 3$, $b = -2$, and $c = -6$:

• Calculate the discriminant: $D = (-2)^2 - 4(3)(-6) = 4 + 72 = 76$

• Then, $x = \frac{2 \pm \sqrt{76}}{6}$ Simplifying gives: $x = \frac{2 \pm 2\sqrt{19}}{6} \Rightarrow x = \frac{1 \pm \sqrt{19}}{3}$

Computing the values, we get:

• $x \approx 1.79$ and $x \approx -1.12$

To solve this equation, we can square both sides to eliminate the square root:

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

Expanding the right-hand side gives:

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

Rearranging leads to:

$x^2 - 4x = 0$

Factoring: $x(x - 4) = 0$

Thus, the solutions are:

• $x = 0$
• $x = 4$

However, we must check if both solutions satisfy the original equation.

Rearranging the inequality gives:

$x^2 - 2x - 3 > 0$

Factoring yields:

$(x - 3)(x + 1) > 0$

Using critical points, $x = -1$ and $x = 3$, we can test intervals:

• For $x < -1$, the expression is positive.
• For $-1 < x < 3$, the expression is negative.
• For $x > 3$, the expression is positive.

Thus, the solution is:

• $x < -1$ or $x > 3$

From the first equation, express $y$: $y = \frac{x + 2}{2}$

Substituting this into the second equation:

$\frac{1}{x} + \frac{x + 2}{2} = 1$

Multiplying through by $2x$ simplifies to:

$2 + x(x + 2) = 2x$ $x^2 - x + 2 = 0$

Using the quadratic formula to solve for x: $x = \frac{1 \pm \sqrt{1 - 8}}{2}$

As the discriminant is negative, the solutions do not yield real numbers for $x$. Therefore, we revert to the relation:

• If $y = 2$, then $x = -1$.

Rearranging and recognizing the powers gives:

$2^{m+4} + 2^{n} = 3^{n} - 3$

Testing integers, we check $m = 3$ and $n = 0$, yielding a trivial equation: $2^{7} + 2^{0} = 3^{0} - 3$

• Thus, $m = 3$ and $n = 0$, leading to: m + n = 3.