Solve for \( x \) to IVO decimal places where necessary: 1.1.1 \( 3 x+\frac{1}{x}=4 \) 1.1.2 \( 5 x(x-3)=2 \) 1.1.3 \( x^{2}-2 x>3 \)

Let's solve each of the equations one by one. 1.1.1 \( 3x + \frac{1}{x} = 4 \) To eliminate the fraction, multiply through by \( x \) (assuming \( x \neq 0 \)): \[ 3x^2 + 1 = 4x \] Rearranging: \[ 3x^2 - 4x + 1 = 0 \] Now, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3, b = -4, c = 1 \): \[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} \] \[ x = \frac{4 \pm \sqrt{16 - 12}}{6} \] \[ x = \frac{4 \pm 2}{6} \] This gives us two solutions: \[ x_1 = \frac{6}{6} = 1 \quad \text{and} \quad x_2 = \frac{2}{6} = \frac{1}{3} \] 1.1.2 \( 5x(x - 3) = 2 \) First, expand the left side: \[ 5x^2 - 15x - 2 = 0 \] Now apply the quadratic formula where \( a = 5, b = -15, c = -2 \): \[ x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4 \cdot 5 \cdot (-2)}}{2 \cdot 5} \] \[ x = \frac{15 \pm \sqrt{225 + 40}}{10} \] \[ x = \frac{15 \pm \sqrt{265}}{10} \] Evaluating \( \sqrt{265} \approx 16.28 \): \[ x \approx \frac{15 \pm 16.28}{10} \] So, we have: \[ x_1 \approx \frac{31.28}{10} = 3.128 \quad \text{and} \quad x_2 \approx \frac{-1.28}{10} = -0.128 \] 1.1.3 \( x^{2} - 2x > 3 \) Rearranging gives us: \[ x^2 - 2x - 3 > 0 \] Factoring the left side: \[ (x - 3)(x + 1) > 0 \] To find the intervals where the product is positive, consider the critical points \( x = 3 \) and \( x = -1 \). We test intervals: - For \( x < -1 \): Both factors are negative, product is positive. - For \( -1 < x < 3 \): One factor is negative, product is negative. - For \( x > 3 \): Both factors are positive, product is positive. Thus, the solution set is: \[ (-\infty, -1) \cup (3, \infty) \]