1.1 Solve for x: 1.1.1 $x^2 + 5x - 6 = 0$ 1.1.2 $4x^2 + 3x - 5 = 0$ (correct to TWO decimal places) 1.1.3 $4x^2 - 1 < 0$ 1.1.4 $\left( \sqrt{32+x} \right) \left( \sqrt{32-x} \right) = x$ 1.2 Solve simultaneously for x and y: $y + x = 12$ and $xy = 14 - 3x$ 1.3 Consider the product $1 \times 2 \times 3 \times 4 \times 30$ - NSC Mathematics - Question 1 - 2019 - Paper 1

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 4$, $b = 3$, and $c = -5$:

$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 4 \cdot (-5)}}{2 \cdot 4} \implies x = \frac{-3 \pm \sqrt{9 + 80}}{8} = \frac{-3 \pm \sqrt{89}}{8}$

Calculating the values:

1. $x = \frac{-3 + \sqrt{89}}{8} \approx 0.80$ (to two decimal places)
2. $x = \frac{-3 - \sqrt{89}}{8} \approx -1.55$ (to two decimal places)

Thus, the values are approximately $x \approx 0.80$ and $x \approx -1.55$.

To solve the inequality, we first rewrite it as:

$4x^2 < 1$

Dividing both sides by 4:

$x^2 < \frac{1}{4}$

Taking the square root of both sides gives:

$-\frac{1}{2} < x < \frac{1}{2}$

The solution set is therefore $x \in \left(-\frac{1}{2}, \frac{1}{2}\right)$.

First, we can square both sides to eliminate the square roots:

$\left( \sqrt{32+x} \right) \left( \sqrt{32-x} \right) = x \implies \sqrt{32+x} \cdot \sqrt{32-x} = x.$

This simplifies to:

$\sqrt{(32+x)(32-x)} = x \implies 32^2 - x^2 = x^2 \implies 1024 - x^2 = x^2.$

Rearranging gives:

$1024 = 2x^2 \implies x^2 = 512 \implies x = \sqrt{512} = 16 \implies x = 4 \text{ (considering only positive solutions)}.$

From the first equation, we can express y in terms of x:

$y = 12 - x.$

Substituting this into the second equation:

$x(12 - x) = 14 - 3x \implies 12x - x^2 = 14 - 3x \implies x^2 - 15x + 14 = 0.$

Factoring this gives:

$(x - 14)(x - 1) = 0$

Thus, $x = 14$ or $x = 1$. Now substituting back to find y:

1. If $x = 14$, then $y = 12 - 14 = -2$.
2. If $x = 1$, then $y = 12 - 1 = 11$.

The solutions for (x, y) pairs are thus $(14, -2)$ and $(1, 11)$.

To determine the largest value of k such that $3^k$ divides the product:

1. Factor each term in the product:

   • $1 = 1$
   • $2 = 2$
   • $3 = 3^1$
   • $4 = 2^2$
   • $30 = 2 \cdot 3 \cdot 5$

2. Now count the total factors of 3:

   • From $3$: $1$ factor
   • From $30$: $1$ factor

Thus, the total number of factors of 3 in the product is $1 + 1 = 2$. Therefore, $k = 2$.