1.3. Given that \( 9 x^{2}-12 p x=-4 p^{2} \) : Determine the value(s) of \( p \) for which the equation will have equal roots. (4) [29] QUESTION 2 2.1. Simplify the following expressions fully, WITHOUT the use of a calculator: 2.1.1. \( 8^{\frac{2}{3}} \) (2) \( 2.1 .2 \quad \frac{2}{1+\sqrt{2}}-\frac{8}{\sqrt{8}} \) \( 2.1 .3 \frac{\sqrt{27}+\sqrt{49-1}-\sqrt{75}}{(\sqrt{3})^{3}} \) (4) 2. Solve for \( x \) if: \[ \left(2^{x+1}\right)^{4} \cdot 2^{1-2 x}=1728 \]

To have equal roots in a quadratic equation, the discriminant must be zero. For the equation \( 9x^2 - 12px + 4p^2 = 0 \), the discriminant \( D \) is given by \( D = b^2 - 4ac \). Here, \( a=9 \), \( b=-12p \), and \( c=4p^2 \). Setting the discriminant to zero, you would find that \( (-12p)^2 - 4 \cdot 9 \cdot 4p^2 = 0 \), leading to the equation \( 144p^2 - 144p^2 = 0 \). Thus, any value of \( p \) will yield equal roots due to this balance. When simplifying \( 8^{\frac{2}{3}} \), rewrite 8 as \( 2^3 \). Thus, \( 8^{\frac{2}{3}} = (2^3)^{\frac{2}{3}} = 2^{3 \cdot \frac{2}{3}} = 2^2 = 4 \). This method of breaking down exponents helps make complex expressions more manageable, and it’s a handy trick in the math toolkit! To tackle the expression \( \frac{2}{1+\sqrt{2}}-\frac{8}{\sqrt{8}} \), rationalize the first term by multiplying the numerator and denominator by \( 1 - \sqrt{2} \) to get \( \frac{2(1 - \sqrt{2})}{-1} = 2\sqrt{2} - 2 \). For the second term, simplify \( \frac{8}{\sqrt{8}} \) to \( 4\). Therefore, combining these results gives you \( 2\sqrt{2} - 6 \). When solving \( (2^{x+1})^{4} \cdot 2^{1-2x}=1728 \), rewrite it as \( 2^{4(x+1)+1-2x} = 1728 \). Note that \( 1728 = 12^3 = (2^2 \cdot 3)^3 = 2^6 \cdot 3^3 \) leads to the equation \( 4x + 4 + 1 - 2x = 6 \). From here, simplify to find \( 2x + 5 = 6 \), so \( x = \frac{1}{2} \). Solving these exponentials can sometimes feel like cracking a code!