Solve for \( x \) in each of the following 1.1.1 \( 3 x^{2}-5 x-1=0 \) (leaveyo 1.1.2 \( \quad x^{2}-6 x+8=0 \) 1.1.3 \( \quad 4 x-2 x^{2}<0 \) 1.4 \( \quad 2^{1 x+1}+2^{3 x}=12 \) \( 1.5 \quad \sqrt{x-1}+3=x-4 \)

To solve these equations for \( x \), let's tackle them one by one in a fun and engaging manner! 1.1.1: For the equation \( 3x^{2} - 5x - 1 = 0 \), we can use the quadratic formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \). Here, \( a = 3 \), \( b = -5 \), and \( c = -1 \). Plugging these values into the formula gives us \( x = \frac{{5 \pm \sqrt{{(-5)^2 - 4(3)(-1)}}}}{{2(3)}} = \frac{{5 \pm \sqrt{{25 + 12}}}}{{6}} = \frac{{5 \pm \sqrt{37}}}{6} \). So, the solutions are \( x \approx 1.54 \) or \( x \approx -0.11 \). 1.1.2: For \( x^{2} - 6x + 8 = 0 \), we can factor it as \( (x - 2)(x - 4) = 0 \). Setting each factor to zero gives us \( x - 2 = 0 \) or \( x - 4 = 0 \), which leads to the solutions \( x = 2 \) and \( x = 4 \). Two simple solutions, straight from the roots! 1.1.3: The inequality \( 4x - 2x^{2} < 0 \) can be rewritten as \( -2x^{2} + 4x < 0 \) or \( 2x^{2} - 4x > 0 \). Factoring out gives us \( 2x(x - 2) > 0 \). Testing intervals, the solution comes from \( x < 0 \) or \( x > 2 \). 1.4: For \( 2^{x+1} + 2^{3x} = 12 \), we can rewrite it as \( 2 \cdot 2^x + 2^{3x} = 12 \). Let \( y = 2^x \), we get \( 2y + y^3 = 12 \). Rearranging gives us the polynomial \( y^3 + 2y - 12 = 0 \). Solving via trial or the Rational Root Theorem, we can find \( y = 2 \). Thus, \( 2^x = 2 \Rightarrow x = 1 \). 1.5: For \( \sqrt{x - 1} + 3 = x - 4 \), let's isolate the square root: \( \sqrt{x - 1} = x - 7 \). Squaring both sides gives \( x - 1 = (x - 7)^2 \). Expanding and rearranging leads to \( x^2 - 15x + 48 = 0 \). Factoring gives \( (x - 3)(x - 12) = 0 \), so \( x = 3 \) or \( x = 12 \). However, don't forget to check for extraneous solutions by plugging back into the original equation! With these step-by-step solutions, you’re equipped to tackle similar problems with confidence! Happy solving!