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

Ask by Burton Gray. in South Africa
Mar 15,2025

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Answer

1.1.2: \(\frac{1}{125}\) 1.2: \(\frac{3\sqrt{x} + 2x\sqrt{x}}{2x}\) 1.3.1: \(x \approx -1.37\) or \(x \approx 0.68\) 1.3.2: \(x \approx -0.63\) or \(x = 0.5\) 1.3.3: \(x < -1\) or \(x > 3\) 1.3.4: \(x = 7\) 1.4: \(x = 3\), \(y = 0\)

Solution

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: \((x, y) = (3, 0)\)

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The Deep Dive

Let's dive into the world of exponents and equations with a sprinkle of history and application! Exponents have an ancient history, dating back to the Babylonians who used forms of exponentiation in their number systems around 2000 BC! They may not have had the same symbols we use today, but concepts like powers were already in existence, primarily for practical calculations in trade, astronomy, and agriculture, paving the way for modern algebra. As for real-world application, simplifying expressions can ease complex calculations, like those in finance. When figuring out compounded interest or investment growth, simplifying exponential terms can help quickly assess growth over time. Using these skills, you can become your own financial wizard, making sense of numbers that might initially seem overwhelming!

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