Solve for \( x \) in cach of the following: 11.1 \( 3 x^{2}-5 x-1 \) - 0 - leave your answer coirect to t Wo decimat \( 1.12 \quad x^{2}-6 x+8=0 \) \( 1.1 .3 \quad 4 x-2 x^{2}-0 \) \( 1.1 .4 \quad 2^{3 x+1}+2^{3 x}=12 \) \( 1.1 .5 \sqrt{x-1}+3=x-4 \)

**Problem 1** Solve \[ 3x^2 - 5x - 1 = 0 \] using the quadratic formula: \[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \] with \(a=3\), \(b=-5\), and \(c=-1\). 1. Compute the discriminant: \[ b^2-4ac = (-5)^2-4(3)(-1)=25+12=37. \] 2. Substitute into the formula: \[ x=\frac{-(-5)\pm\sqrt{37}}{2\cdot 3}=\frac{5\pm\sqrt{37}}{6}. \] 3. Approximating to two decimals: - For the positive root: \[ x\approx\frac{5+6.08}{6}=\frac{11.08}{6}\approx1.85. \] - For the negative root: \[ x\approx\frac{5-6.08}{6}=\frac{-1.08}{6}\approx-0.18. \] **Problem 2** Solve \[ x^2 -6x +8=0. \] Factor the quadratic: \[ x^2-6x+8=(x-2)(x-4)=0. \] Thus, \[ x=2\quad\text{or}\quad x=4. \] **Problem 3** Solve \[ 4x-2x^2=0. \] Factor the expression: \[ 2x(2-x)=0. \] Set each factor equal to zero: 1. \(2x=0\) yields \(x=0\). 2. \(2-x=0\) yields \(x=2\). Therefore, \(x=0\) or \(x=2\). **Problem 4** Solve \[ 2^{3x+1}+2^{3x}=12. \] Notice that \[ 2^{3x+1}=2\cdot2^{3x}. \] Then the equation becomes: \[ 2\cdot2^{3x}+2^{3x}=3\cdot2^{3x}=12. \] Solve for \(2^{3x}\): \[ 2^{3x}=\frac{12}{3}=4. \] Since \(4=2^2\), we have: \[ 2^{3x}=2^2, \] which implies \[ 3x=2 \quad\Longrightarrow\quad x=\frac{2}{3}. \] **Problem 5** Solve \[ \sqrt{x-1}+3=x-4. \] 1. Isolate the square root: \[ \sqrt{x-1}=x-4-3=x-7. \] Note that for the square root to be defined, we require: \[ x-1\ge0 \quad\Longrightarrow\quad x\ge1, \] and since the square root is nonnegative, also: \[ x-7\ge0\quad\Longrightarrow\quad x\ge7. \] 2. Square both sides: \[ (\sqrt{x-1})^2=(x-7)^2 \quad\Longrightarrow\quad x-1=x^2-14x+49. \] 3. Rearrange to form a quadratic equation: \[ 0=x^2-14x+49-(x-1)=x^2-15x+50. \] 4. Solve the quadratic using the quadratic formula: \[ x=\frac{15\pm\sqrt{15^2-4\cdot1\cdot50}}{2}=\frac{15\pm\sqrt{225-200}}{2}=\frac{15\pm\sqrt{25}}{2}. \] So, \[ x=\frac{15\pm5}{2}. \] This gives: - \(x=\frac{15+5}{2}=\frac{20}{2}=10\), - \(x=\frac{15-5}{2}=\frac{10}{2}=5\). 5. Check the solutions against the restriction \(x\ge7\). Since \(5<7\) it must be rejected. 6. Verify \(x=10\) by substituting back: \[ \sqrt{10-1}+3= \sqrt{9}+3= 3+3=6\quad\text{and}\quad10-4=6. \] The equation holds true. Thus, \(x=10\).