3.1 Simplify the following without the use of a calculator: 3.1.1 \( \sqrt{8x^{27}} \) 3.1.2 \( 9^{1} \cdot 4^{1} \cdot 6^{-2n} \) 3.1.3 \( \sqrt{2 - \sqrt{-4}} - \sqrt{4k} \) 3.2 Given: \( \frac{\log 72 - \log 2}{\log 6} \) 3.2.1 Write the following as a single logarithm: \( \log 72 - \log 2 \) 3.2.2 Hence, simplify without using a calculator: \( \frac{\log 72 - \log 2}{\log 6} \) 3.3 Solve for \( x: 5x^{2} - 5x = 600 \) 3.4 Given the complex numbers: \( r_{1} = 2 + 3i \) and \( r_{2} = i \) 3.4.1 Write down the conjugate of \( r_{2} \) 3.4.2 Hence, simplify \( \frac{r_{1}}{r_{2}} \) Show ALL steps - NSC Technical Mathematics - Question 3 - 2024 - Paper 1

To simplify ( 9^{1} \cdot 4^{1} \cdot 6^{-2n} ), we use the prime factorization and properties of exponents:

( 9 = 3^{2}, ) ( 4 = 2^{2}, ) ( 6 = 2 \cdot 3. )

Thus: $9^{1} \cdot 4^{1} \cdot 6^{-2n} = 3^{2} \cdot 2^{2} \cdot (2 \cdot 3)^{-2n} \quad = 3^{2} \cdot 2^{2} \cdot 2^{-2n} \cdot 3^{-2n}$

Combining terms yields: $= 3^{2 - 2n} \cdot 2^{2 - 2n}$

We first simplify ( \sqrt{2 - \sqrt{-4}} ). Since ( \sqrt{-4} = 2i ), we have:

( 2 - \sqrt{-4} = 2 - 2i. ) Using the expression in polar form, we find: $\sqrt{2 - 2i} \text{ can be solved using polar coordinates.}$

Next, we simplify ( \sqrt{4k} ) to ( 2\sqrt{k} ). Thus:

Final form is ( \sqrt{2 - 2i} - 2\sqrt{k} ).

To write ( \log 72 - \log 2 ) as a single logarithm, apply the logarithmic property that states ( \log a - \log b = \log \left( \frac{a}{b} \right) ). Therefore:

$\log 72 - \log 2 = \log \left( \frac{72}{2} \right) = \log 36.$

Using the result from step 3.2.1:

We can substitute to find:

$\frac{\log 36}{\log 6} = \log_{6} 36 = 2. \quad \text{(since } 36 = 6^{2})$

To solve the equation, first rearrange: $5x^{2} - 5x - 600 = 0.$

We can divide through by 5: $x^{2} - x - 120 = 0.$

Now, use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-120)}}{2 \cdot 1}$

Simplifying further, $= \frac{1 \pm \sqrt{1 + 480}}{2} = \frac{1 \pm \sqrt{481}}{2}$

The conjugate of a complex number ( r = a + bi ) is ( a - bi ). Thus, the conjugate of ( r_{2} = i ) is given by:

$\text{Conjugate of } r_{2} = -i.$

Given ( r_{1} = 2 + 3i ) and ( r_{2} = i ), we have:

$\frac{r_{1}}{r_{2}} = \frac{2 + 3i}{i}.$

To simplify, multiply the numerator and the denominator by the conjugate of the denominator:

$= \frac{(2 + 3i)(-i)}{i \cdot (-i)} = \frac{-2i - 3(-1)}{-1} = 2i + 3.$

Thus, the simplified expression is: $\frac{r_{1}}{r_{2}} = 3 + 2i.$

To express ( a + bi = -i -14 ) in standard form, we find:

Equating real and imaginary parts, we find: ( a = -14 ) and ( b = -1. )

Thus the values are:

( a = -14, b = -1. )