1.1 Los op vir $x$: 1.1.1 $x^2 + 5x - 6 = 0$ 1.1.2 $4x^2 + 3x - 5 = 0$ (korrek tot TWEE desimale plekke) 1.1.3 $4x^2 - 1 < 0$ 1.1.4 $\left( \sqrt{32} + x \right) \left( \sqrt{32} - x \right) = x$ 1.2 Los gelijktidig op vir $x$ en $y$: $y + x = 12$ en $xy = 14 - 3x$ - NSC Mathematics - Question 1 - 2019 - Paper 1

We can apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 4$, $b = 3$, and $c = -5$. Substituting these values, we have:

$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 4 \cdot (-5)}}{2 \cdot 4} = \frac{-3 \pm \sqrt{9 + 80}}{8} = \frac{-3 \pm \sqrt{89}}{8}$ Calculating the two roots yields:

$x_1 \approx -0.55 \quad \text{and} \quad x_2 \approx 0.8$

To solve this inequality, we first factor the left-hand side:

$4x^2 - 1 < 0 \implies (2x - 1)(2x + 1) < 0$ Next, we find the critical points:

$2x - 1 = 0 \implies x = 0.5 \quad \text{and} \quad 2x + 1 = 0 \implies x = -0.5$ We check intervals around the critical points:

• For $x < -0.5$, the product is positive.
• For $-0.5 < x < 0.5$, the product is negative.
• For $x > 0.5$, the product is positive. Thus, the solution is:

$-0.5 < x < 0.5$

To solve this equation, we can first expand the left-hand side:

$\sqrt{32}^2 - x^2 = x \implies 32 - x^2 = x$ Rearranging gives:

$x^2 + x - 32 = 0$ Now we can factor or apply the quadratic formula: Using the quadratic formula:

$x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-32)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 128}}{2} = \frac{-1 \pm \sqrt{129}}{2}$ Thus, we find:

$x = 4 \quad \text{(selecting the positive root)}$

Starting with the first equation, express $y$:

$y = 12 - x$ Substituting into the second equation:

$(12 - x)x = 14 - 3x$ Expanding gives:

$12x - x^2 = 14 - 3x \implies x^2 - 15x + 14 = 0$ Factoring, we have:

$(x - 14)(x - 1) = 0$ Thus, $x = 14$ or $x = 1$. Then, substituting back to find $y$:

For $x = 14$: $y = -2$ (not valid) For $x = 1$: $y = 11$. Hence, valid values are $x = 1, y = 11$.

Calculating the product gives:

$1 \times 2 \times 3 \times 4 \times 30 = 240$ To find the highest power of $3$ that divides $240$, we determine its prime factorization:

$240 = 2^4 \times 3^1 \times 5^1$ Thus, the greatest value of $k$ such that $3^k$ is a factor is:

$k = 1$