QUESTION 1 \( \begin{array}{l}1.1 \text { Simplify: } \\ \text { 1.1.2 } \frac{25^{x+1} \cdot 25^{x-2}}{125^{x+1} \cdot 5^{x-2}} \\ 1.2 \text { Rationalise the denominator } \\ \frac{6}{4 \sqrt{x}}+\sqrt{x} \\ 1.3 \text { Solve for } x \\ 1.3 .13 x^{2}+9 x=12 \\ 1.3 .22 x^{2}+3 x-7=0 \\ 1.3 .3(3-x)(1+x)<0 \\ 1.3 .4 \sqrt{x+18}+2=x \\ \text { (round off your answer to TWO decimal places) } \\ 1.4 \text { Solve for } x \text { and } y \\ 2 x^{2}+y(4 x-1)=15 \\ x+y=3\end{array} \)

Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(13x^{2}+9x-12=0\) - step1: Solve using the quadratic formula: \(x=\frac{-9\pm \sqrt{9^{2}-4\times 13\left(-12\right)}}{2\times 13}\) - step2: Simplify the expression: \(x=\frac{-9\pm \sqrt{9^{2}-4\times 13\left(-12\right)}}{26}\) - step3: Simplify the expression: \(x=\frac{-9\pm \sqrt{705}}{26}\) - step4: Separate into possible cases: \(\begin{align}&x=\frac{-9+\sqrt{705}}{26}\\&x=\frac{-9-\sqrt{705}}{26}\end{align}\) - step5: Rewrite the fraction: \(\begin{align}&x=\frac{-9+\sqrt{705}}{26}\\&x=-\frac{9+\sqrt{705}}{26}\end{align}\) - step6: Rewrite: \(x_{1}=-\frac{9+\sqrt{705}}{26},x_{2}=\frac{-9+\sqrt{705}}{26}\) Calculate or simplify the expression \( \frac{25^{x+1} \cdot 25^{x-2}}{125^{x+1} \cdot 5^{x-2}} \). Simplify the expression by following steps: - step0: Solution: \(\frac{25^{x+1}\times 25^{x-2}}{125^{x+1}\times 5^{x-2}}\) - step1: Multiply: \(\frac{25^{2x-1}}{125^{x+1}\times 5^{x-2}}\) - step2: Multiply: \(\frac{25^{2x-1}}{5^{4x+1}}\) - step3: Factor the expression: \(\frac{5^{4x-2}}{5^{4x+1}}\) - step4: Divide the numbers: \(\frac{1}{5^{4x+1-\left(4x-2\right)}}\) - step5: Subtract the terms: \(\frac{1}{5^{3}}\) - step6: Evaluate the power: \(\frac{1}{125}\) Calculate or simplify the expression \( \frac{6}{4 \sqrt{x}}+\sqrt{x} \). Simplify the expression by following steps: - step0: Solution: \(\frac{6}{4\sqrt{x}}+\sqrt{x}\) - step1: Reduce the fraction: \(\frac{3}{2\sqrt{x}}+\sqrt{x}\) - step2: Reduce fractions to a common denominator: \(\frac{3}{2\sqrt{x}}+\frac{\sqrt{x}\times 2\sqrt{x}}{2\sqrt{x}}\) - step3: Transform the expression: \(\frac{3+\sqrt{x}\times 2\sqrt{x}}{2\sqrt{x}}\) - step4: Multiply the terms: \(\frac{3+2x}{2\sqrt{x}}\) - step5: Simplify: \(\frac{3\sqrt{x}+2x\sqrt{x}}{2x}\) Solve the equation \( 3(3-x)(1+x)<0 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(3\left(3-x\right)\left(1+x\right)<0\) - step1: Rewrite the expression: \(3\left(3-x\right)\left(1+x\right)=0\) - step2: Elimination the left coefficient: \(\left(3-x\right)\left(1+x\right)=0\) - step3: Separate into possible cases: \(\begin{align}&3-x=0\\&1+x=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=3\\&x=-1\end{align}\) - step5: Determine the test intervals: \(\begin{align}&x<-1\\&-13\end{align}\) - step6: Choose a value: \(\begin{align}&x_{1}=-2\\&x_{2}=1\\&x_{3}=4\end{align}\) - step7: Test the chosen value: \(\begin{align}&x<-1\textrm{ }\textrm{is the solution}\\&-13\textrm{ }\textrm{is the solution}\end{align}\) - step8: The final solution is \(x \in \left(-\infty,-1\right)\cup \left(3,+\infty\right):\) \(x \in \left(-\infty,-1\right)\cup \left(3,+\infty\right)\) Solve the equation \( \sqrt{x+18}+2=x \). Solve the equation by following steps: - step0: Solve for \(x\): \(\sqrt{x+18}+2=x\) - step1: Find the domain: \(\sqrt{x+18}+2=x ,x\geq -18\) - step2: Move the constant to the right-hand side: \(\sqrt{x+18}=x-2\) - step3: Evaluate: \(\sqrt{x+18}=x-2,x-2\geq 0\) - step4: Evaluate: \(\sqrt{x+18}=x-2,x\geq 2\) - step5: Solve the equation: \(\begin{align}&x=7\\&x=-2\end{align},x\geq 2\) - step6: Find the intersection: \(x=7\) - step7: Check if the solution is in the defined range: \(x=7,x\geq -18\) - step8: Find the intersection: \(x=7\) - step9: Check the solution: \(x=7\) Solve the system of equations \( 2 x^{2}+y(4 x-1)=15;x+y=3 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2x^{2}+y\left(4x-1\right)=15\\x+y=3\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}2x^{2}+y\left(4x-1\right)=15\\x=3-y\end{array}\right.\) - step2: Substitute the value of \(x:\) \(2\left(3-y\right)^{2}+y\left(4\left(3-y\right)-1\right)=15\) - step3: Simplify: \(18-y-2y^{2}=15\) - step4: Move the expression to the left side: \(18-y-2y^{2}-15=0\) - step5: Subtract the numbers: \(3-y-2y^{2}=0\) - step6: Factor the expression: \(\left(1-y\right)\left(3+2y\right)=0\) - step7: Separate into possible cases: \(\begin{align}&1-y=0\\&3+2y=0\end{align}\) - step8: Solve the equation: \(\begin{align}&y=1\\&y=-\frac{3}{2}\end{align}\) - step9: Calculate: \(y=1\cup y=-\frac{3}{2}\) - step10: Rearrange the terms: \(\left\{ \begin{array}{l}x=3-y\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=3-y\\y=-\frac{3}{2}\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=2\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{9}{2}\\y=-\frac{3}{2}\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=2\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{9}{2}\\y=-\frac{3}{2}\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(2,1\right)\cup \left(x,y\right) = \left(\frac{9}{2},-\frac{3}{2}\right)\) Solve the equation \( 22 x^{2}+3 x-7=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(22x^{2}+3x-7=0\) - step1: Factor the expression: \(\left(2x-1\right)\left(11x+7\right)=0\) - step2: Separate into possible cases: \(\begin{align}&2x-1=0\\&11x+7=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=\frac{1}{2}\\&x=-\frac{7}{11}\end{align}\) - step4: Rewrite: \(x_{1}=-\frac{7}{11},x_{2}=\frac{1}{2}\) Let's solve each part of the problem step by step. ### 1.1 Simplify #### 1.1.2 We need to simplify the expression: \[ \frac{25^{x+1} \cdot 25^{x-2}}{125^{x+1} \cdot 5^{x-2}} \] **Step 1:** Rewrite the bases: - \( 25 = 5^2 \) - \( 125 = 5^3 \) **Step 2:** Substitute: \[ \frac{(5^2)^{x+1} \cdot (5^2)^{x-2}}{(5^3)^{x+1} \cdot 5^{x-2}} = \frac{5^{2(x+1)} \cdot 5^{2(x-2)}}{5^{3(x+1)} \cdot 5^{x-2}} \] **Step 3:** Combine the exponents: \[ = \frac{5^{2x + 2 + 2x - 4}}{5^{3x + 3 + x - 2}} = \frac{5^{4x - 2}}{5^{4x + 1}} = 5^{(4x - 2) - (4x + 1)} = 5^{-3} = \frac{1}{125} \] So, the simplified result is: \[ \frac{1}{125} \] ### 1.2 Rationalise the denominator We need to rationalise: \[ \frac{6}{4 \sqrt{x}} + \sqrt{x} \] **Step 1:** Find a common denominator: \[ = \frac{6 + 4x}{4\sqrt{x}} \] **Step 2:** Rationalise the denominator: \[ = \frac{(6 + 4x) \cdot \sqrt{x}}{4x} = \frac{6\sqrt{x} + 4x\sqrt{x}}{4x} \] So, the rationalised expression is: \[ \frac{3\sqrt{x} + 2x\sqrt{x}}{2x} \] ### 1.3 Solve for \( x \) #### 1.3.1 Solve: \[ 13x^{2} + 9x - 12 = 0 \] The solutions are: \[ x_{1} \approx -1.37, \quad x_{2} \approx 0.68 \] #### 1.3.2 Solve: \[ 22x^{2} + 3x - 7 = 0 \] The solutions are: \[ x_{1} \approx -0.63, \quad x_{2} = 0.5 \] #### 1.3.3 Solve the inequality: \[ 3(3-x)(1+x) < 0 \] The solution set is: \[ x \in (-\infty, -1) \cup (3, +\infty) \] #### 1.3.4 Solve: \[ \sqrt{x+18} + 2 = x \] The solution is: \[ x = 7 \] ### 1.4 Solve for \( x \) and \( y \) We need to solve the system: \[ 2x^{2} + y(4x - 1) = 15 \] \[ x + y = 3 \] **Step 1:** Substitute \( y = 3 - x \) into the first equation: \[ 2x^{2} + (3 - x)(4x - 1) = 15 \] **Step 2:** Expand and simplify: \[ 2x^{2} + 12x - 4x^{2} - 3 = 15 \] \[ -2x^{2} + 12x - 18 = 0 \] \[ x^{2} - 6x + 9 = 0 \] \[ (x - 3)^{2} = 0 \implies x = 3 \] **Step 3:** Substitute back to find \( y \): \[ y = 3 - 3 = 0 \] So, the solution is: \[ (x, y) = (3, 0) \] ### Summary of Results 1.1.2: \(\frac{1}{125}\) 1.2: \(\frac{3\sqrt{x} + 2x\sqrt{x}}{2x}\) 1.3.1: \(x_{1} \approx -1.37, x_{2} \approx 0.68\) 1.3.2: \(x_{1} \approx -0.63, x_{2} = 0.5\) 1.3.3: \(x \in (-\infty, -1) \cup (3, +\infty)\) 1.3.4: \(x = 7\) 1.4: \((x, y) = (3, 0)\)