Los op vir $x$: 1.1.1 $x^2 + 2x - 15 = 0$ (3) 1.1.2 $5x^2 - x - 9 = 0$ (Los jou antwoord korrek tot TWEE desimale syfers.) (4) 1.1.3 $x^2 \, \leq \, 3x$ (4) 1.2 Gegee: $\frac{a + 64}{a} = 16$ (3) 1.2.1 Los op vir $a$ - NSC Mathematics - Question 1 - 2022 - Paper 1

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 5$, $b = -1$, and $c = -9$:

1. Calculate the discriminant: $\Delta = (-1)^2 - 4 \cdot 5 \cdot (-9) = 1 + 180 = 181$

2. Substitute into the formula: $x = \frac{-(-1) \pm \sqrt{181}}{2 \cdot 5} = \frac{1 \pm \sqrt{181}}{10}$

Calculating the two potential values for $x$ gives us:

1. $x = \frac{1 + \sqrt{181}}{10} \approx 1.45$
2. $x = \frac{1 - \sqrt{181}}{10} \approx -1.25$

Therefore, the solutions for $x$ are approximately $x = 1.45$ and $x = -1.25$.

Rearranging the equation gives:

$x^2 - 3x \leq 0$

We can factor it:

$(x)(x - 3)\leq 0$

The critical points are $x = 0$ and $x = 3$. To find the intervals where the inequality holds, we check the signs:

• For $x < 0$: both factors are negative, so the product is positive.
• For $0 < x < 3$: the first factor is positive and the second is negative, hence the product is negative.
• For $x = 3$: the product is $0$.
• For $x > 3$: both factors are positive, so the product is positive.

Thus, the solution set is:

$0 \leq x \leq 3$

To solve for $a$, we start by multiplying both sides by $a$ to eliminate the fraction:

$a + 64 = 16a$

Rearranging gives:

$64 = 16a - a$

Thus:

$64 = 15a$

Finally, dividing both sides by $15$:

$a = \frac{64}{15}$

Rewriting $16$ as a power of $2$ gives:

$16 = 2^4$

The equation becomes:

$2^x + 2^{6x} = 2^4$

For simplification, we can express $2^{6x}$ as $(2^x)^6$:

Letting $y = 2^x$, we have:

$y + y^6 = 2^4$

To solve for $y$, this can be rewritten as:

$y^6 + y - 16 = 0$

Finding the solution for $y$ gives:

$y = 2 \implies 2^x = 2 \implies x = 1$

First, simplify the expression in the square root:

$2^{1002} + 2^{1006} = 2^{1002}(1 + 2^4) = 2^{1002}(1 + 16) = 17 \cdot 2^{1002}$

So, we now have:

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

From $2x - y = 2$, we can express $y$ in terms of $x$:

$y = 2x - 2$

Substituting this expression into the second equation $\frac{1}{x} - 3y = 1$ gives:

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

Simplifying leads to:

$\frac{1}{x} - 6x + 6 = 1$

Rearranging gives:

$\frac{1}{x} - 6x + 5 = 0$

Multiplying through by $x$:

$1 - 6x^2 + 5x = 0$

Thus it becomes:

$6x^2 - 5x - 1 = 0$

Using the quadratic formula:

$x = \frac{5 \pm \sqrt{(-5)^2 - 4(6)(-1)}}{2(6)}$

This gives us the values for $x$. After finding $x$, substitute back to find $y$. The values will be: $x = \frac{1}{6}$ or $x = 1$.