Solve for \( x \) in each of the following \( 1.1 .1 \quad 3 x^{2}-5 x-1=0 \) (leave \( y \) \( 1.1 .2 \quad x^{2}-6 x+8=0 \) 1.1 .3 \( 4 x-2 x^{2}<0 \)

Solve the quadratic equation by following steps: - step0: Solve by factoring: \(x^{2}-6x+8=0\) - step1: Factor the expression: \(\left(x-4\right)\left(x-2\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x-4=0\\&x-2=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=4\\&x=2\end{align}\) - step4: Rewrite: \(x_{1}=2,x_{2}=4\) Solve the equation \( 3x^{2}-5x-1=0 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(3x^{2}-5x-1=0\) - step1: Solve using the quadratic formula: \(x=\frac{5\pm \sqrt{\left(-5\right)^{2}-4\times 3\left(-1\right)}}{2\times 3}\) - step2: Simplify the expression: \(x=\frac{5\pm \sqrt{\left(-5\right)^{2}-4\times 3\left(-1\right)}}{6}\) - step3: Simplify the expression: \(x=\frac{5\pm \sqrt{37}}{6}\) - step4: Separate into possible cases: \(\begin{align}&x=\frac{5+\sqrt{37}}{6}\\&x=\frac{5-\sqrt{37}}{6}\end{align}\) - step5: Rewrite: \(x_{1}=\frac{5-\sqrt{37}}{6},x_{2}=\frac{5+\sqrt{37}}{6}\) Solve the equation \( 4x-2x^{2}<0 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(4x-2x^{2}<0\) - step1: Rewrite the expression: \(4x-2x^{2}=0\) - step2: Factor the expression: \(2x\left(2-x\right)=0\) - step3: Separate into possible cases: \(\begin{align}&2x=0\\&2-x=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=0\\&x=2\end{align}\) - step5: Determine the test intervals: \(\begin{align}&x<0\\&02\end{align}\) - step6: Choose a value: \(\begin{align}&x_{1}=-1\\&x_{2}=1\\&x_{3}=3\end{align}\) - step7: Test the chosen value: \(\begin{align}&x<0\textrm{ }\textrm{is the solution}\\&02\textrm{ }\textrm{is the solution}\end{align}\) - step8: The final solution is \(x \in \left(-\infty,0\right)\cup \left(2,+\infty\right):\) \(x \in \left(-\infty,0\right)\cup \left(2,+\infty\right)\) Let's solve each of the equations step by step. ### 1.1.1: Solve \( 3x^{2}-5x-1=0 \) Using the quadratic formula, we find the solutions for \( x \): \[ x_{1} = \frac{5 - \sqrt{37}}{6}, \quad x_{2} = \frac{5 + \sqrt{37}}{6} \] ### 1.1.2: Solve \( x^{2}-6x+8=0 \) This is a simple quadratic equation. The solutions are: \[ x_{1} = 2, \quad x_{2} = 4 \] ### 1.1.3: Solve \( 4x-2x^{2}<0 \) To solve the inequality, we first rewrite it: \[ -2x^{2} + 4x < 0 \quad \Rightarrow \quad 2x^{2} - 4x > 0 \quad \Rightarrow \quad 2x(x - 2) > 0 \] The critical points are \( x = 0 \) and \( x = 2 \). Testing intervals around these points, we find: \[ x \in (-\infty, 0) \cup (2, +\infty) \] ### Summary of Solutions: 1. \( 3x^{2}-5x-1=0 \) gives \( x_{1} = \frac{5 - \sqrt{37}}{6}, x_{2} = \frac{5 + \sqrt{37}}{6} \) 2. \( x^{2}-6x+8=0 \) gives \( x_{1} = 2, x_{2} = 4 \) 3. \( 4x-2x^{2}<0 \) gives \( x \in (-\infty, 0) \cup (2, +\infty) \)