1.1 Solve for $x$: 1.1.1 $x^{2}+2x-15=0$ 1.1.2 $5x^{2}-x-9=0$ (Leave your answer correct to TWO decimal places.) 1.1.3 $x^{2} \\leq 3x$ 1.2 Given: $\frac{a + 64}{a} = 16$ 1.2.1 Solve for $a$ - NSC Mathematics - Question 1 - 2022 - Paper 1

Using the quadratic formula:

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

For the equation, $a=5$, $b=-1$, $c=-9$:

First, calculate the discriminant: $b^{2}-4ac = (-1)^{2} - 4 \cdot 5 \cdot (-9) = 1 + 180 = 181$

Now plug values into the formula:

$x = \frac{-(-1) \pm \sqrt{181}}{2 \cdot 5} = \frac{1 \pm \sqrt{181}}{10}$

Calculating the roots gives:

$x \approx \frac{1 + 13.45}{10} \approx 1.45\text{ (to 2 decimal places)}$ $x \approx \frac{1 - 13.45}{10} \approx -1.25$

Rearrange the inequality:

$x^{2} - 3x \leq 0$

Factoring gives: $(x)(x-3) \leq 0$

Using a sign chart, the critical points are $x=0$ and $x=3$. Testing intervals, we find:

The solution to the inequality is: $0 \leq x \leq 3$

Start with the equation:

$\frac{a + 64}{a} = 16$

Cross-multiply to eliminate the fraction:

$a + 64 = 16a$

Rearranging leads to: $64 = 15a \implies a = \frac{64}{15} \approx 4.27$

Simplify the equation:

Since $16 = 2^{4}$, we have:

$2^{x} + 64 = 16 \implies 2^{x} = 16 - 64 = -48$

Since $2^{x}$ cannot be negative, we conclude that no solutions exist for this part.

Factor out $2^{1002}$ in the numerator:

$= \sqrt{\frac{2^{1002}(1 + 2^{4})}{17(2)^{99}}}$

This simplifies to:

$= \sqrt{\frac{2^{1002} \cdot 17}{17 \cdot 2^{99}}} \implies \sqrt{2^{3}} = \sqrt{8} = 2\sqrt{2}$

We have the equations:

$2x - y = 2 \\ (1)$ $\frac{1}{x} - 3y = 1 \\ (2)$

From (1), express $y$ in terms of $x$: $y = 2x - 2$

Substituting into (2): $\frac{1}{x} - 3(2x - 2) = 1$

Simplify the equation to find: $\frac{1}{x} - 6x + 6 = 1 \\ \implies 6x^{2} - 6x + 1 = 0$

Using the quadratic formula to find $x$, followed by substituting back to find $y$.