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

To solve the quadratic equation $5x^2 - x - 9 = 0$, we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Here, $a = 5$, $b = -1$, and $c = -9$. Calculating the discriminant:

$b^2 - 4ac = (-1)^2 - 4(5)(-9) = 1 + 180 = 181$

Now substituting into the formula:

$x = \frac{1 \pm \sqrt{181}}{10}$

Calculating the values, we find:

• $x \approx 1.45$ (to two decimal places)
• $x \approx -1.25$.

First, rearranging gives us:
$x^2 - 3x \leq 0$
Factoring yields:
$(x)(x - 3) \leq 0$

To find the intervals, we identify critical points at $x = 0$ and $x = 3$. Analyzing the sign of the product in each interval:

• For $x < 0$, the product is positive.
• For $0 \leq x \leq 3$, the product is non-positive.
• For $x > 3$, the product is positive.

Thus, the solution is:

$0 \leq x \leq 3$

Starting with the equation:

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

Multiplying both sides by $a$:

$a + 64 = 16a$

Rearranging gives:

$64 = 15a$

Thus, solving for $a$ results in:

$a = \frac{64}{15} \approx 4.27$

From part 1.2.1, we established $a$. Now we begin:

$2^x + 2^{x-6} = 16$

We can express $16$ as $2^4$, enabling us to combine terms:

$2^x + \frac{2^x}{64} = 2^4$

Factoring out $2^x$ gives:

$2^x \left(1 + \frac{1}{64}\right) = 2^4$

Simplifying: $2^x \cdot \frac{65}{64} = 2^4$

Thus, $2^x = 2^4 \cdot \frac{64}{65}$

The resultant exponent gives us the solution for x: $x \approx 3$.

We begin by simplifying the expression:

$\sqrt{\frac{2^{1002} + 2^{1006}}{17(2)^{98}}}$

By factoring out $2^{1002}$ from the numerator:

$= \sqrt{\frac{2^{1002}(1 + 2^4)}{17(2)^{98}}}$

This simplifies to:

$= \sqrt{\frac{2^{1002} \cdot 17}{17 \cdot 2^{98}}}$ = \sqrt{2^4} = 2$$.

We start with the given equations:

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

From equation 1, we can express $y$:

$y = 2x - 2$

Now substituting this into equation 2:

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

Expanding yields:

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

Rearranging gives:

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

Multiplying through by $x$ (assuming $x \neq 0$):

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

Rearranging yields:
$6x^2 - 5x - 1 = 0$

Using the quadratic formula: $x = \frac{5 \pm \sqrt{(-5)^2 - 4(6)(-1)}}{2(6)}$

Calculating gives result for $x$ values and substituting back to find corresponding $y$ values.