Solve for \( x \) in each of the followin 1.1.1 \( 3 x^{2}-5 x-1=0 \) (leave? 1.1.2 \( x^{2}-6 x+8=0 \) \( 1.3 \quad 4 x-2 x^{2}<0 \) 1.4 \( 2^{1.0+1}+2^{3 x}=12 \) 1.5 \( \sqrt{x-1}+3=x-4 \)

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 \( \sqrt{x-1}+3=x-4 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\sqrt{x-1}+3=x-4\) - step1: Find the domain: \(\sqrt{x-1}+3=x-4,x\geq 1\) - step2: Move the constant to the right side: \(\sqrt{x-1}=x-4-3\) - step3: Calculate: \(\sqrt{x-1}=x-7\) - step4: Evaluate: \(\sqrt{x-1}=x-7,x-7\geq 0\) - step5: Evaluate: \(\sqrt{x-1}=x-7,x\geq 7\) - step6: Solve the equation: \(\begin{align}&x=10\\&x=5\end{align},x\geq 7\) - step7: Find the intersection: \(x=10\) - step8: Check if the solution is in the defined range: \(x=10,x\geq 1\) - step9: Find the intersection: \(x=10\) - step10: Check the solution: \(x=10\) Solve the equation \( 2^{1.0+1}+2^{3x}=12 \). Solve the equation by following steps: - step0: Solve for \(x\): \(2^{1+1}+2^{3x}=12\) - step1: Simplify: \(4+2^{3x}=12\) - step2: Move the constant to the right side: \(2^{3x}=12-4\) - step3: Subtract the numbers: \(2^{3x}=8\) - step4: Rewrite in exponential form: \(2^{3x}=2^{3}\) - step5: Set the exponents equal: \(3x=3\) - step6: Divide both sides: \(\frac{3x}{3}=\frac{3}{3}\) - step7: Divide the numbers: \(x=1\) 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 \(x \in \left(-\infty,0\right)\cup \left(2,+\infty\right):\) \(x \in \left(-\infty,0\right)\cup \left(2,+\infty\right)\) Here are the solutions for each of the equations: ### 1.1.1 \( 3x^{2}-5x-1=0 \) The solutions are: \[ x_{1} = \frac{5 - \sqrt{37}}{6}, \quad x_{2} = \frac{5 + \sqrt{37}}{6} \] ### 1.1.2 \( x^{2}-6x+8=0 \) The solutions are: \[ x_{1} = 2, \quad x_{2} = 4 \] ### 1.3 \( 4x-2x^{2}<0 \) The solution set is: \[ x \in (-\infty, 0) \cup (2, +\infty) \] ### 1.4 \( 2^{1.0+1}+2^{3x}=12 \) The solution is: \[ x = 1 \] ### 1.5 \( \sqrt{x-1}+3=x-4 \) The solution is: \[ x = 10 \] If you have any further questions or need additional assistance, feel free to ask!