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: \[ \left(2^{x+1}\right)^{4} \cdot 2^{1-2 x}=1728 \]

Solve the equation by following steps: - step0: Solve for \(p\): \(9x^{2}-12px+4p^{2}=0\) - step1: Rewrite the expression: \(9x^{2}-12xp+4p^{2}=0\) - step2: Factor the expression: \(\left(3x-2p\right)^{2}=0\) - step3: Simplify the expression: \(3x-2p=0\) - step4: Move the expression to the right side: \(-2p=0-3x\) - step5: Remove 0: \(-2p=-3x\) - step6: Change the signs: \(2p=3x\) - step7: Divide both sides: \(\frac{2p}{2}=\frac{3x}{2}\) - step8: Divide the numbers: \(p=\frac{3x}{2}\) Solve the equation \( (2^{x+1})^{4} \cdot 2^{1-2x}=1728 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\left(2^{x+1}\right)^{4}\times 2^{1-2x}=1728\) - step1: Simplify: \(2^{2x+5}=1728\) - step2: Take the logarithm of both sides: \(\log_{2}{\left(2^{2x+5}\right)}=\log_{2}{\left(1728\right)}\) - step3: Evaluate the logarithm: \(2x+5=\log_{2}{\left(1728\right)}\) - step4: Move the constant to the right side: \(2x=\log_{2}{\left(1728\right)}-5\) - step5: Simplify: \(2x=3\log_{2}{\left(12\right)}-5\) - step6: Divide both sides: \(\frac{2x}{2}=\frac{3\log_{2}{\left(12\right)}-5}{2}\) - step7: Divide the numbers: \(x=\frac{3\log_{2}{\left(12\right)}-5}{2}\) - step8: Simplify: \(x=\frac{1+3\log_{2}{\left(3\right)}}{2}\) Calculate or simplify the expression \( 8^{\frac{2}{3}} \). Calculate the value by following steps: - step0: Calculate: \(8^{\frac{2}{3}}\) - step1: Rewrite in exponential form: \(\left(2^{3}\right)^{\frac{2}{3}}\) - step2: Multiply the exponents: \(2^{3\times \frac{2}{3}}\) - step3: Multiply the exponents: \(2^{2}\) - step4: Evaluate the power: \(4\) Calculate or simplify the expression \( \frac{2}{1+\sqrt{2}}-\frac{8}{\sqrt{8}} \). Calculate the value by following steps: - step0: Calculate: \(\frac{2}{1+\sqrt{2}}-\frac{8}{\sqrt{8}}\) - step1: Simplify the root: \(\frac{2}{1+\sqrt{2}}-\frac{8}{2\sqrt{2}}\) - step2: Reduce the fraction: \(\frac{2}{1+\sqrt{2}}-\frac{4}{\sqrt{2}}\) - step3: Calculate: \(-2+2\sqrt{2}-\frac{4}{\sqrt{2}}\) - step4: Calculate: \(-2+2\sqrt{2}-2\sqrt{2}\) - step5: Rewrite the expression: \(\left(-1+\sqrt{2}\right)\times 2-\sqrt{2}\times 2\) - step6: Factor the expression: \(\left(-1+\sqrt{2}-\sqrt{2}\right)\times 2\) - step7: Calculate: \(-2\) Calculate or simplify the expression \( \frac{\sqrt{27}+\sqrt{49-1}-\sqrt{75}}{(\sqrt{3})^{3}} \). Calculate the value by following steps: - step0: Calculate: \(\frac{\sqrt{27}+\sqrt{49-1}-\sqrt{75}}{\left(\sqrt{3}\right)^{3}}\) - step1: Subtract the numbers: \(\frac{\sqrt{27}+\sqrt{48}-\sqrt{75}}{\left(\sqrt{3}\right)^{3}}\) - step2: Simplify the root: \(\frac{3\sqrt{3}+\sqrt{48}-\sqrt{75}}{\left(\sqrt{3}\right)^{3}}\) - step3: Simplify the root: \(\frac{3\sqrt{3}+4\sqrt{3}-\sqrt{75}}{\left(\sqrt{3}\right)^{3}}\) - step4: Simplify the root: \(\frac{3\sqrt{3}+4\sqrt{3}-5\sqrt{3}}{\left(\sqrt{3}\right)^{3}}\) - step5: Calculate: \(\frac{2\sqrt{3}}{\left(\sqrt{3}\right)^{3}}\) - step6: Simplify the expression: \(\frac{2\sqrt{3}}{3\sqrt{3}}\) - step7: Reduce the fraction: \(\frac{2}{3}\) Let's solve the problems step by step. ### Problem 1: Determine the value(s) of \( p \) for which the equation will have equal roots. The given equation is: \[ 9x^{2} - 12px + 4p^{2} = 0 \] For a quadratic equation \( ax^2 + bx + c = 0 \) to have equal roots, the discriminant must be zero: \[ D = b^2 - 4ac = 0 \] Here, \( a = 9 \), \( b = -12p \), and \( c = 4p^2 \). Calculating the discriminant: \[ D = (-12p)^2 - 4(9)(4p^2) = 144p^2 - 144p^2 = 0 \] Since the discriminant is already zero, it indicates that the equation will have equal roots for any value of \( p \). However, if we want to express \( p \) in terms of \( x \), we can rearrange the equation: \[ p = \frac{3x}{2} \] ### Problem 2: Simplify the following expressions fully. #### 2.1.1: Simplify \( 8^{\frac{2}{3}} \) Using the property of exponents: \[ 8^{\frac{2}{3}} = (2^3)^{\frac{2}{3}} = 2^{3 \cdot \frac{2}{3}} = 2^2 = 4 \] #### 2.1.2: Simplify \( \frac{2}{1+\sqrt{2}} - \frac{8}{\sqrt{8}} \) First, simplify \( \frac{8}{\sqrt{8}} \): \[ \frac{8}{\sqrt{8}} = \frac{8}{2\sqrt{2}} = \frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \] Now, substitute back: \[ \frac{2}{1+\sqrt{2}} - 2\sqrt{2} \] To simplify \( \frac{2}{1+\sqrt{2}} \), multiply the numerator and denominator by \( 1 - \sqrt{2} \): \[ \frac{2(1 - \sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})} = \frac{2(1 - \sqrt{2})}{1 - 2} = -2(1 - \sqrt{2}) = 2\sqrt{2} - 2 \] Now combine: \[ (2\sqrt{2} - 2) - 2\sqrt{2} = -2 \] #### 2.1.3: Simplify \( \frac{\sqrt{27}+\sqrt{49-1}-\sqrt{75}}{(\sqrt{3})^{3}} \) Calculating each square root: \[ \sqrt{27} = 3\sqrt{3}, \quad \sqrt{49-1} = \sqrt{48} = 4\sqrt{3}, \quad \sqrt{75} = 5\sqrt{3} \] Substituting back: \[ \frac{3\sqrt{3} + 4\sqrt{3} - 5\sqrt{3}}{(\sqrt{3})^{3}} = \frac{(3 + 4 - 5)\sqrt{3}}{3\sqrt{3}} = \frac{2\sqrt{3}}{3\sqrt{3}} = \frac{2}{3} \] ### Problem 3: Solve for \( x \) if: \[ (2^{x+1})^{4} \cdot 2^{1-2x} = 1728 \] First, simplify the left side: \[ (2^{x+1})^{4} = 2^{4(x+1)} = 2^{4x + 4} \] Thus, the equation becomes: \[ 2^{4x + 4} \cdot 2^{1 - 2x} = 1728 \] Combine the exponents: \[ 2^{(4x + 4) + (1 - 2x)} = 1728 \implies 2^{2x + 5} = 1728 \] Now express \( 1728 \) as a power of \( 2 \): \[ 1728 = 2^6 \cdot 3^3 \] Since \( 1728 \) is not a pure power of \( 2 \), we can equate the powers: \[ 2x + 5 = 6 \implies 2x = 1 \implies x = \frac{1}{2} \] ### Summary of Results: 1. \( p = \frac{3x}{2} \) (for equal roots) 2. \( 8^{\frac{2}{3}} = 4 \) 3. \( \frac{2}{1+\sqrt{2}} - \frac{8}{\sqrt{8}} = -2 \) 4. \( \frac{\sqrt{27}+\sqrt{49-1}-\sqrt{75}}{(\sqrt{3})^{3}} = \frac{2}{3} \) 5. \( x = \frac{1}{2} \) (from the exponential equation)