Solve for \( x \), without using a calculator: \( \begin{array}{l}\text { 2.1.1 } \quad 6 x^{2}-x-1=0 \\ 2.1 .2 \quad 3^{2 x-1}=\frac{1}{243}\end{array} \)

To solve the first equation \( 6x^2 - x - 1 = 0 \), we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 6 \), \( b = -1 \), and \( c = -1 \). Plugging in these values: 1. Calculate the discriminant: \[ b^2 - 4ac = (-1)^2 - 4(6)(-1) = 1 + 24 = 25 \] 2. Now use the quadratic formula: \[ x = \frac{-(-1) \pm \sqrt{25}}{2(6)} = \frac{1 \pm 5}{12} \] 3. This gives the solutions: \[ x = \frac{6}{12} = \frac{1}{2} \quad \text{and} \quad x = \frac{-4}{12} = -\frac{1}{3} \] For the second equation \( 3^{2x-1} = \frac{1}{243} \), we start by expressing \( \frac{1}{243} \) in terms of base 3. We know that \( 243 = 3^5 \), so \( \frac{1}{243} = 3^{-5} \). Thus, we set the exponents equal: \[ 2x - 1 = -5 \] Now, solving for \( x \): 1. Add 1 to both sides: \[ 2x = -5 + 1 = -4 \] 2. Divide by 2: \[ x = -2 \] So the solutions are \( x = \frac{1}{2} \) and \( x = -\frac{1}{3} \) for the first equation, and \( x = -2 \) for the second equation.