\( x(x+3)-1=0 \) (Leave your answer in simplest surd form \( 2 x=5 x^{2}+3 \) Consider: \( 5 x-\frac{3}{x}=1 \) (a) Solve for \( x \) correct to two decimal places.

Let’s start by solving the first equation, \( x(x + 3) - 1 = 0 \). Rearranging, we have: \[ x^2 + 3x - 1 = 0 \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1, b = 3, c = -1 \): \[ x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-1)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 4}}{2} = \frac{-3 \pm \sqrt{13}}{2} \] The solutions are: \[ x = \frac{-3 + \sqrt{13}}{2} \quad \text{and} \quad x = \frac{-3 - \sqrt{13}}{2} \] In simplest surd form, we keep these as they are. Now for the second equation \( 2x = 5x^2 + 3 \). Rearranging gives: \[ 5x^2 - 2x + 3 = 0 \] Using the quadratic formula again, here \( a = 5, b = -2, c = 3 \): \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(5)(3)}}{2(5)} = \frac{2 \pm \sqrt{4 - 60}}{10} = \frac{2 \pm \sqrt{-56}}{10} \] This indicates complex solutions, since we have a negative determinant. We can simplify it: \[ \sqrt{-56} = \sqrt{56}i = 2\sqrt{14}i \] Thus the solutions are: \[ x = \frac{2 \pm 2\sqrt{14}i}{10} = \frac{1 \pm \sqrt{14}i}{5} \] Lastly, for the equation \( 5x - \frac{3}{x} = 1 \), we can multiply both sides by \( x \) (assuming \( x \neq 0 \)): \[ 5x^2 - 3 = x \Rightarrow 5x^2 - x - 3 = 0 \] Use the quadratic formula again: \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(5)(-3)}}{2(5)} = \frac{1 \pm \sqrt{1 + 60}}{10} = \frac{1 \pm \sqrt{61}}{10} \] Calculating the decimal approximations gives: 1. \( x \approx 0.86 \) and \( x \approx -0.66 \) for the real solutions. The complete solution set is: - \( x = \frac{-3 + \sqrt{13}}{2} \) (≈ -0.30) - \( x = \frac{-3 - \sqrt{13}}{2} \) (≈ -2.70) - \( x = \frac{1 + \sqrt{61}}{10} \) (≈ 1.79) - \( x = \frac{1 - \sqrt{61}}{10} \) (≈ -0.67) Enjoy your calculations!