Solve the equation 2x - $\dfrac{1}{x} = 7$. Write your answer correct to two decimal places.

Given,

$\Rightarrow 2x - \dfrac{1}{x} = 7 \\[1em] \Rightarrow \dfrac{2x^2 - 1}{x} = 7 \\[1em] \Rightarrow 2x^2 - 1 = 7x \\[1em] \Rightarrow 2x^2 - 7x - 1 = 0$

Comparing 2x² - 7x - 1 = 0 with ax² + bx + c = 0 we get,

a = 2, b = -7 and c = -1.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-7) \pm \sqrt{(-7)^2 - 4.(2).(-1)}}{2(2)} \\[1em] = \dfrac{7 \pm \sqrt{49 + 8}}{4} \\[1em] = \dfrac{7 \pm \sqrt{57}}{4} \\[1em] = \dfrac{7 \pm 7.55}{4} \\[1em] = \dfrac{7 + 7.55}{4} \text{ or } \dfrac{7 - 7.55}{4} \\[1em] = \dfrac{14.55}{4} \text{ or } \dfrac{-0.55}{4} \\[1em] = 3.64 \text{ or } -0.14$

Hence, x = 3.64 or -0.14