OUESTION I (NOV 2008) 11 Solve for \( x \), rounded off to TWO decumal places where necessary \( 111 \quad x^{2}=5 x-4 \) \( 112 \quad x(3-x)=-3 \) 113 3- \( x<2 x^{2} \) (5) 1.2 Determine the values of \( x \) and \( y \) if they satisfy both the following equations simultancously: \[ \begin{array}{l} 2 x+y=3 \\ x^{2}+y+x=y^{2} \end{array} \] (8 3 Given \( x=999999999999 \), determine the exact value of \( \frac{x^{2}-4}{x-2} \) Show ALL your calculations. 4 Explain why the equation \( \frac{x^{4}+1}{x^{4}}=\frac{1}{2} \) has no real roots.

Calculate the value by following steps: - step0: Calculate: \(\frac{\left(999999999999^{2}-4\right)}{\left(999999999999-2\right)}\) - step1: Remove the parentheses: \(\frac{999999999999^{2}-4}{999999999999-2}\) - step2: Subtract the numbers: \(\frac{999999999999^{2}-4}{999999999997}\) Solve the equation \( \frac{x^{4}+1}{x^{4}}=\frac{1}{2} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{x^{4}+1}{x^{4}}=\frac{1}{2}\) - step1: Cross multiply: \(\left(x^{4}+1\right)\times 2=x^{4}\) - step2: Simplify the equation: \(2\left(x^{4}+1\right)=x^{4}\) - step3: Expand the expression: \(2x^{4}+2=x^{4}\) - step4: Move the expression to the left side: \(2x^{4}+2-x^{4}=0\) - step5: Subtract the terms: \(x^{4}+2=0\) - step6: Move the constant to the right side: \(x^{4}=0-2\) - step7: Remove 0: \(x^{4}=-2\) - step8: The statement is false for any value of \(x:\) \(x \notin \mathbb{R}\) Solve the equation \( x(3-x)=-3 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(x\left(3-x\right)=-3\) - step1: Expand the expression: \(3x-x^{2}=-3\) - step2: Move the expression to the left side: \(3x-x^{2}+3=0\) - step3: Rewrite in standard form: \(-x^{2}+3x+3=0\) - step4: Multiply both sides: \(x^{2}-3x-3=0\) - step5: Solve using the quadratic formula: \(x=\frac{3\pm \sqrt{\left(-3\right)^{2}-4\left(-3\right)}}{2}\) - step6: Simplify the expression: \(x=\frac{3\pm \sqrt{21}}{2}\) - step7: Separate into possible cases: \(\begin{align}&x=\frac{3+\sqrt{21}}{2}\\&x=\frac{3-\sqrt{21}}{2}\end{align}\) - step8: Rewrite: \(x_{1}=\frac{3-\sqrt{21}}{2},x_{2}=\frac{3+\sqrt{21}}{2}\) Solve the equation \( 111 x^{2}=5 x-4 \). Solve the equation(The complex numbers system) by following steps: - step0: Solve using the quadratic formula in the complex numbers system: \(111x^{2}=5x-4\) - step1: Move the expression to the left side: \(111x^{2}-5x+4=0\) - step2: Solve using the quadratic formula: \(x=\frac{5\pm \sqrt{\left(-5\right)^{2}-4\times 111\times 4}}{2\times 111}\) - step3: Simplify the expression: \(x=\frac{5\pm \sqrt{\left(-5\right)^{2}-4\times 111\times 4}}{222}\) - step4: Simplify the expression: \(x=\frac{5\pm \sqrt{-1751}}{222}\) - step5: Simplify the expression: \(x=\frac{5\pm \sqrt{1751}\times i}{222}\) - step6: Separate into possible cases: \(\begin{align}&x=\frac{5+\sqrt{1751}\times i}{222}\\&x=\frac{5-\sqrt{1751}\times i}{222}\end{align}\) - step7: Simplify the expression: \(\begin{align}&x=\frac{5}{222}+\frac{\sqrt{1751}}{222}i\\&x=\frac{5-\sqrt{1751}\times i}{222}\end{align}\) - step8: Simplify the expression: \(\begin{align}&x=\frac{5}{222}+\frac{\sqrt{1751}}{222}i\\&x=\frac{5}{222}-\frac{\sqrt{1751}}{222}i\end{align}\) - step9: Rewrite: \(x_{1}=\frac{5}{222}-\frac{\sqrt{1751}}{222}i,x_{2}=\frac{5}{222}+\frac{\sqrt{1751}}{222}i\) - step10: Remove the complex number(s): \(\textrm{No real solution}\) Solve the equation \( 3-x<2 x^{2} \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(3-x<2x^{2}\) - step1: Move the expression to the left side: \(3-x-2x^{2}<0\) - step2: Rewrite the expression: \(3-x-2x^{2}=0\) - step3: Factor the expression: \(\left(1-x\right)\left(3+2x\right)=0\) - step4: Separate into possible cases: \(\begin{align}&1-x=0\\&3+2x=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=1\\&x=-\frac{3}{2}\end{align}\) - step6: Determine the test intervals: \(\begin{align}&x<-\frac{3}{2}\\&-\frac{3}{2}1\end{align}\) - step7: Choose a value: \(\begin{align}&x_{1}=-3\\&x_{2}=0\\&x_{3}=2\end{align}\) - step8: Test the chosen value: \(\begin{align}&x<-\frac{3}{2}\textrm{ }\textrm{is the solution}\\&-\frac{3}{2}1\textrm{ }\textrm{is the solution}\end{align}\) - step9: The final solution is \(x \in \left(-\infty,-\frac{3}{2}\right)\cup \left(1,+\infty\right):\) \(x \in \left(-\infty,-\frac{3}{2}\right)\cup \left(1,+\infty\right)\) Solve the system of equations \( 2 x+y=3;x^{2}+y+x=y^{2} \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2x+y=3\\x^{2}+y+x=y^{2}\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}y=3-2x\\x^{2}+y+x=y^{2}\end{array}\right.\) - step2: Substitute the value of \(y:\) \(x^{2}+3-2x+x=\left(3-2x\right)^{2}\) - step3: Add the terms: \(x^{2}+3-x=\left(3-2x\right)^{2}\) - step4: Expand the expression: \(x^{2}+3-x=9-12x+4x^{2}\) - step5: Move the expression to the left side: \(x^{2}+3-x-\left(9-12x+4x^{2}\right)=0\) - step6: Calculate: \(-3x^{2}-6+11x=0\) - step7: Factor the expression: \(\left(-x+3\right)\left(3x-2\right)=0\) - step8: Separate into possible cases: \(\begin{align}&-x+3=0\\&3x-2=0\end{align}\) - step9: Solve the equation: \(\begin{align}&x=3\\&x=\frac{2}{3}\end{align}\) - step10: Calculate: \(x=3\cup x=\frac{2}{3}\) - step11: Rearrange the terms: \(\left\{ \begin{array}{l}x=3\\y=3-2x\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{2}{3}\\y=3-2x\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=3\\y=-3\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{2}{3}\\y=\frac{5}{3}\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=3\\y=-3\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{2}{3}\\y=\frac{5}{3}\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(3,-3\right)\cup \left(x,y\right) = \left(\frac{2}{3},\frac{5}{3}\right)\) Let's solve the problems step by step. ### Problem 1: Solve for \( x \) 1. **Equation:** \( 111 x^{2} = 5 x - 4 \) Rearranging gives us: \[ 111 x^{2} - 5 x + 4 = 0 \] This quadratic equation has no real solutions. 2. **Equation:** \( x(3-x) = -3 \) Rearranging gives us: \[ x^2 - 3x - 3 = 0 \] Using the quadratic formula, we find: \[ x_{1} = \frac{3 - \sqrt{21}}{2}, \quad x_{2} = \frac{3 + \sqrt{21}}{2} \] 3. **Inequality:** \( 3 - x < 2 x^{2} \) Rearranging gives us: \[ 2 x^{2} + x - 3 > 0 \] The solution set is: \[ x \in (-\infty, -\frac{3}{2}) \cup (1, +\infty) \] ### Problem 2: Determine values of \( x \) and \( y \) Given the equations: \[ \begin{align*} 2x + y &= 3 \\ x^{2} + y + x &= y^{2} \end{align*} \] We can solve this system of equations. The solutions are: \[ (x, y) = (3, -3) \cup (x, y) = \left(\frac{2}{3}, \frac{5}{3}\right) \] In decimal form, this is: \[ (x, y) = (3, -3) \cup (x, y) = (0.67, 1.67) \] ### Problem 3: Calculate \( \frac{x^{2}-4}{x-2} \) Given \( x = 999999999999 \): \[ \frac{x^{2}-4}{x-2} = \frac{999999999999^{2}-4}{999999999997} \] This expression simplifies to: \[ 999999999999 + 2 = 1000000000001 \] ### Problem 4: Explain why \( \frac{x^{4}+1}{x^{4}}=\frac{1}{2} \) has no real roots Rearranging gives: \[ x^{4} + 1 = \frac{1}{2} x^{4} \] This simplifies to: \[ \frac{1}{2} x^{4} + 1 = 0 \] Since \( x^{4} \) is always non-negative, \( \frac{1}{2} x^{4} + 1 \) is always positive. Therefore, there are no real solutions to this equation. ### Summary of Results 1. **First equation:** No real solution. 2. **Second equation:** \( x_{1} = \frac{3 - \sqrt{21}}{2}, x_{2} = \frac{3 + \sqrt{21}}{2} \). 3. **Third inequality:** \( x \in (-\infty, -\frac{3}{2}) \cup (1, +\infty) \). 4. **Values of \( x \) and \( y \):** \( (3, -3) \cup \left(\frac{2}{3}, \frac{5}{3}\right) \). 5. **Value of expression:** \( 1000000000001 \). 6. **Fourth equation:** No real roots.